A different Application of the preceding Example. 1422. The model (fig. 580.) is an arch similar to that of the preceding example, having a story above it formed by two walls, whose height is 100, and the whole covered by a timber roof. The object of the investigation is to ascertain what change may be made in the thickness of the piers which are strengthened in their resistance by the additional weight upon them. 1423. The simplest method of proceeding is to consider the upper walls as prolongations of the piers. 1424. In the model the walls were made of plaster, and their weight was thus reduced to of what they would have been if of the stone used for the models hitherto described. The roof weighed 12 ounces. We shall therefore have that 100, which in stone would have represented the weight of the walls, from the difference in weight of the plaster, reduced to 75. In respect of the roof, which weighed 12 ounces, having found by experiment that it was equal to an area of 576 lines of the stone, both being reduced to equal thicknesses, we have 12 ounces, equal to an area of 13.82 whose half 6'91 must be added to that of the vertical efforts represented by b in and Changing these terms into and the formula becomes bb aa hh aa' The height of the piers or a in the formula-183 +75 = 258. d, the difference between the height of the pier and the arm of the lever, will=75. Substituting these values in the formula, we shall have x= √265.86-77·28 +31·22—5·58=9·15. In the model a thickness of 11 lines was found sufficient to resist the thrust, and taking the root of double the thrust the result is 13 lines. 1425. By the geometrical method, given in the last, taking from the result 17 lines, there found, the value of, that is, 5·58, the remainder 113 a, the height of the pier, 198.00. The area KFCL of the upper part of the arch will be 1223.10, from which subtracting that of the triangle FKG, which is 590-82, the remainder 832-28 being multiplied by 30.73 and divided by 32.7 makes the effort of this part 782.44. 1429. The area of the lower part is 697.95, to which adding the triangle FKG=390-82, we have 1088 77, which multiplied by 23.27 and divided by 52.15, gives 485-82 for its effort. The expression of the thrust, represented by p in the formula, a b, representing the sum of the efforts of the semi-arch, will be b 1762-8 a= 198 =8.9 and bb Substituting these values in the formula, we have the equation x= √593-24-100·64+79-21-8-9-15 01. By taking double the square root of the thrust the result is 23-91, a thickness evidently too great, because the sum of the vertical efforts, which are therein neglected, is considerable. 1430. The geometrical method gives 19 lines. actual experiment was 16 lines. The least thickness of the piers from 1431. Rondelet gives a proof of the method by means of the centres of gravity, as in some of the preceding examples, from which he obtains a result of only 13-26 for the thickness of the piers. Consideration of an Arch whose Voussoirs increase towards the Springing. 1432. The model (fig. 582.) has an extrados of segmental form not concentric with its intrados, so that its thickness increases from the crown to the springing. The opening is the same as before, namely, 9 inches, or 108 lines. The thickness at the vertex is 4 lines, towards the middle of the haunches 7 lines, and at the springing 14 lines. The centre of the line of the extrados is one sixth part of the chord AO below the centre of the intrados; so that The radius DN=68-05 KL=38.18 The arc BKKC=42-43 1433. The area KHDC of the upper part of the arch is 258.75, that of the lower part BAHK 486 5; hence the effort of the upper part is represented by the expression 42.43 258 75 x 38-18 =232.47. 1434. The half segment A Be being supposed to be united to the pier; BeHK, whose area is 178, is the only part that can balance the = 66.24. The difference 178 x 15.82 upper effort; its expression will be 42-43 PSR Fig. 582. of the two efforts 166-23 will be the expression of the thrust represented by p in the formula b, which indicates the vertical effort of the half arch repre These values being substituted in the formula, will give r = √332-46+ 80.39 +15.56-3.95-16.74 lines. 1435. The smallest thickness of pier that would support the arch in the model was 17 lines. 1436. With the geometrical method, instead of the double of CD, make Bh double the mean thickness HK, and Bn equal to mL, and on nh as a diameter describe the semicircumference cutting OB produced in E; then EB-18 lines will be the thickness sought. 1437. If the pier is continued up to the point e where the thickness of the arch is disengaged from the pier, the height of the pier represented in the formula by a will be 151-5 instead of 120, and the difference b, instead of being will be only =277-46. 745 26 × 54 436-75 x 54 85 1438. d, expressed by Ie, will be 6.5, all the other values remaining the same as in the preceding article, the equation is x= √332.46-571+4-2-16.21. 1439. Using the method by means of the centres of gravity, Rondelet found the result for the thickness of the piers to be 15.84. So that there is no great variation in the different results. 1440. In the preceding examples arches have been considered rather as arcades standing on piers than as vaults supported by walls of a certain length. We are now about to consider them in this last respect, and as serving to cover the space enclosed by the walls. In respect of cylindrical arches supported by parallel walls, it is manifest that the resistance they present has no relation to their length; for if we suppose the length of the vault divided into an infinite number of pieces, as C, D, E, &c. (fig. 584. No. 2.), we shall find for each of these pieces the same thickness of pier, so that all the piers together would form a wall of the same thickness. For this reason the surfaces only of the arches and piers have been hitherto considered, that is, as profiles or sections of an arch of any given length. Consequently it may be said that the thickness of wall found for the profile in the section of an arch would serve for the arch continued in length infinitely, supposing such walls isolated and not terminated or rather filled by other walls at their ends. When cylindrical walls are terminated by walls at their extremities, after the manner of gable ends, it is not difficult to imagine that the less distant these walls are the more they add stability to those of the arch. In this case may be applied a rule which we shall hereafter mention more at length under the following section on Walls. 1441. If in any of the examples (fig. 582. for instance) PR be produced indefinitely to the right, and from R on the line so produced the length of the wall supporting the arch be set out, and if from the extremity of such line another be drawn,. as TB produced through B, indefinitely towards a, and Ba be made equal to the thickness of the pier first found, a vertical line let fall from a will determine the thickness sought. When arches are connected with these cross walls, the effect of the thrust may be much diminished if they are not very distant. If there be any openings in the walls, double the length of them must be added to that of the wall as well as of any that may be introduced in the gable wall. 1442. Fig. 583. represents the mode in which an arch fails when the piers are not of sufficient strength to resist the thrust: they open on the lower part of the summit at DM and on the upper part of the haunches at HN; from which we may infer that the thrust of an arch may be destroyed by cramping the under side of the voussoirs near the summit and the upper side of those towards the middle of the haunches; and this method is greatly preferable to chains or iron bars on the extrados, because these have no effect in preventing a failure on the underside. at the springing will not prevent failure in arches whose voussoirs are of equal depth but that too small, inasmuch Chains n that takes place at the haunches, like a hoop loaded when its ends are fixed. The most advantageous position for a chain to oppose the effort of an arch is to let it pass through the point K where the efforts meet. PC is the tangent before failure, and O the centre; R being the inner point of the pier. OF COMPOUND VAULTING. L M F H 1443. M. Frezier, in speaking of the thrust of this sort of arches, proposes, in order to find the thickness of the piers which will support them, to find by the ordinary manner the thickness suitable to each part of the cylindrical arch BN, BK (No. 3. fig. 584.) by which the groin is formed, making BE the thickness suitable to the arch BN, and BF that which the arch BK requires; the pier BEHF would thus be able to resist the thrust of the quarter arch OKBN. According to this method we should find the bay of a groined arch 9 inches opening would not require piers more than 21 lines square and 120 lines high; but experience proves that a similar arch will scarcely stand with H piers 44 lines square, the area of whose bases are four times greater than that proposed by M. Frezier. Method for groined Vaulting. 1444. The model in this case (see the last figure) is 9 inches in the opening, voussoirs equally thick, being 9 lines, standing upon four piers 10 inches or 120 lines high. No. 1. Fig. 584. E 1445. The groin is formed by two eylindrical arches of the same diameter crossing at right angles, as represented in No. 3. of the figure. The four portions of the vault being similar, the calculation for one pier will be sufficient. 1446. On the profile No. 1. of the figure describe the mean circumference TKG, draw the tangents FT and FG, and the secant FO and the horizontal line IKL. Draw the vertical Bi, and NG and KI on the plan (No. 3.) equal to KL 1447. In the foregoing examples for arches and cylindrical vaulting there has been no necessity to consider more than the surface of the profiles, which are constantly the same throughout their length; but the species of vault of which we are now treating being composed of triangular gores whose profile changes at every point, we shall be obliged to use the cubes instead of the areas of squares, and to substitute surfaces for lines. Thus in viewing the triangular part KBO, the sum of the horizontal efforts of the upper part of this portion of the vault, represented in the profile by KL, will be represented in plan by the trapezium KILO. 1448. The sum of those of the lower part iK in the profile is represented in plan by BIL. The thrust is expressed by the difference of the area of the trapezium and of the triangle multiplied by the thickness of the vault; thus, KB and KO of the plan being 54, the superficies of the triangle BKO will be 54 x 27=1458; the part BK of the plan being equal to IL, and Bt to iK of the profile =12, the area of the triangle BIL, indicating the sum of the horizontal efforts of the upper part, will be 12×6=79 1449. We obtain the area of the trapezium KILO by subtracting that of the small triangle BIL from the greater triangle BKO, that is, 79 from 1458; the remainder 1378 gives the horizontal effort of the upper part; lastly, subtracting 79 from 1378 the remainder 1298 will be the expression of the thrust whose value is found by multiplying 1298 by 9=116832, which is the p of the formula, Letting a always stand for the height, and d for TI of the profile, the arm of the lever of the thrust will, as before, be a + d, and its algebraic expression be pa + pd. 1450. The pier resists this effort by its cube multiplied by the arm of its lever. If the lines KB and OB of the triangle BKO, (which represents the projection of that part of the vault for which we are calculating) be produced, it will be seen that the base of the pier to resist the thrust will be represented by the opposite triangle BHF, which is rectangular and isosceles ; therefore, letting a represent its side BF, the area of the triangle will be expressed by the arz height of the pier being a, its cube will be ". The arm of the lever of this pier will be determined by the distance of the vertical let fall from its centre of gravity on the line HF=3, which gives for the pier's resistance arxx 1451. This resistance will be increased by the vertical effort of each part of the vault multiplied by the arm of its lever. That of the upper part will be expressed by its cube multiplied by the vertical KM, and the product divided by the mean arc KG. The cube of this part will be equal to the mean area; that is, the arc KG multiplied by the thickness of the vault. 1452. To obtain the mean area, multiply KG less KM by the length GO taken on the plan. The length of the arc KG being 46 and KM 174, we shall have KG-KM=289; GO being 54, the mean area will be 289 × 54=1558. This area multiplied by 9, the thickness of the vault, makes the cube of the upper part 140244, which multiplied by KM=174 and divided by the arc KG=46, makes 5226 the value of the vertical effort of the part of the arch m in the formula; and the arm of its lever is IK + iH. 1453. IK being =c and iH=x, its expression will be mx + mc. The vertical effort of the lower part will be represented by its cube multiplied by TI, and the product divided by the length of the arc TK. = ar3 This cube will be found by multiplying the mean area by the thickness of the vault. The area being equal to the arc TK-TI× GO, that is, 46–413⁄4× 54=250% for the mean area and 250 × 9=22563 for the cube of the lower part of the vault. This cube 22563 × 41 multiplied by TI and divided by the arc TK gives 20283 for the value of 6 the vertical effort of the part n of the formula. And it is to be observed, that this effort acting against the point B, the arm BF of the lever will be ≈ and its expression nx. 1454. Bringing together all these algebraic values we obtain the equation pa+pd: +mx + mc + nx; and making m+n, which multiplies x=b, we have pa+pd= mc. Transferring me to the other side of the equation, we have pa + pd-mc= Lastly, multiplying all the terms of the equation by for the purpose of eliminating æ3, we shall have instead of the preceding formula 6p + 6pd-6mc =x3+ which is an equation of the third degree, whose second term is wanting. For more easily resolving this equation, let us find the value of 6p+6pd-6mc and that of, by which a is multiplied in the second part of the equation. a p being 116832, 6p will be 6pd-6mc_2361537 a 120 a 65 6 6bx a = 6 6 + bx + 6pd-6mc =19679, and 6p + a = 6 =897794, which we will call g, Thus call f; so that instead of the equation 6p+ 6pd-6mc ≈= √448894+ √2015073623 +1767902 + 448894 – 2/2015073623 +1767902 = 2/448894 +44909+ 2/448894-449092, from which extracting the cube roots, we have x=44-21=42 for the length BF of one of the sides of the triangular pier BAF; the other FA may be determined by the production of the diagonal or line of groin OB. The part of the pier answering to the part of the vault BNO is determined by drawing from the points B and A the parallels BM and MA to FA and FB. These two triangles will form a square base, each of whose sides will be 42 lines, answering to one quarter of the vault KBNO; thus, to resist the thrust of the vault, four piers, each 42 lines thick, are necessary. 1455. The above result corresponds in a singular manner with the experiments which were made by Rondelet, from which he deduced a thickness of 43 lines. In his investigation of the example by means of the centres of gravity 40.53 lines was the result. Our limits prevent further consideration by other examples: we will merely therefore observe, that |