Abbildungen der Seite
PDF

the method above given seems to be a safe guide to the architect. In the case of oblong arches, the results must be obtained for each side. 1456. In the case of groinings composed of many bays, the chief care necessary is in the external piers, which will require especially to be of sufficient thickness. Those in the middle, being counterbalanced all round, have only to bear the weights of their respective arches, for which purpose they must have a proportional area and be of such stone as the weight will not crush. But it ought to be recollected that in good construction the area of the points of support should be so distributed as to establish for each a sufficient strength, because a single weak point will often endanger the whole fabric. 1457. In practice, a readier method will be wanting than that which has been just discussed; we therefore subjoin one which agrees well enough with theory and experiment, and it is as follows. Let ABCD (fig. 585. No. 1.) be the space to be covered by a

groined vault supported in the centre by the pier E. Dividing each side into two equal parts, draw the lines HI, FG crossing each other in the centre K*, K*, and the diagonals AE, EB, EC, ED and HF, HG, IF, IG crossing each other in the points K, K, K”, K". In No. 2 draw the pier, its half height to the level of the springing, which half height transfer from E to K and divide EK into twelve parts. One of these parts will be a half diagonal of the pier. For the intermediate piers H, F, I, G, after finding the diagonals of the half piers, produce them outwards to double their projection within, so that altogether their thickness may be once and a half their width. For the angular piers this method will give an area of base 14 times greater, which will enable them to resist the thrust they have to sustain. 1458. When the width of the space to be vaulted is to be divided into three bays, and that of the middle is required to be raised above those of the other two, as in the case of churches, with side aisles, the bases of the points of support may be determined in two * ...That most used, which is borrowed from the Gothic examples, is to give to the areasof the bases of the points of support merely the extent necessary to bear the load they are to receive, by throwing the strain of the thrust upon the external piers by means of

[graphic]
[graphic]

flying buttresses, and giving to their points of support a position and surface of base capable of effectual resistance.

1459. The most simple method derived from the principles of the theory for the first case is as follows:

Having laid down the plan of the two bays which fall upon the same pier (fig. 586. No. 1.), take one half of the sum of the two semi-diagonals AD, AE, to which add one

half of the height of the point of support, and taking a twelfth part of the whole as a radius, describe a circle, and it will indicate the surface sought of the base of the point of support. If it be not circular it must be circumscribed with the form that may be required, so as rather to increase than diminish its solidity. For the exterior point of support B let a rectangle be formed, having for its width the side of a square inscribed in the preceding circle, and in length double. 1460. Above the roofs of the sides a flying buttress may be carried up, whose pier may be raised on that below, set back one sixth from the exterior face and sloped as much on the interior. The line of summit or tangent of this flying buttress, which should be of the single arc of a circle, will be determined by the chord of the arc of the upper part of the vault produced indefinitely. To find the centre, draw the chord GH (No. 2.), on the middle of which raise a perpendicular, which will cut the horizontal line GF in the point I, which will be the centre of the arc. These raking arches may be connected by - E e 2

[graphic]

other return arches, which may bear a floor above with a support, upon which a passage round the building may be made, and this may be concealed by an attic order outside. 1461. In the second case, the base of a pier must be found capable of resisting the effort of the great middle vault of the nave, by taking as the height of its pier the rise from its springing above the side vaults No. 3., and laying the half of this height from B to H on the plan No. 2. Then having divided IH into twelve equal parts, make IA equal to one of them and AF equal to two. The rectangle made upon the diagonal FI shows the area of the interior pier, to which are to be added, to the right and left, projections to receive the arches of the sides. The length FD is to be divided into six equal parts, whereof two are for the projection of the pilaster or interior half column, upon which the entablature is profiled, three for the thickness of the wall, and one for the pilaster on the side aisles, whose prolongation will form a counterfort above the lower sides. 1462. For the external pier B, as before, one half the height to the springing must be transferred from EG, and of BG from B to L.; lastly, #, from B to K : the rectangle formed upon the diagonal KL is equal to the area of the pier. We must add, as for that in front, the projections to receive the arches or windows, as shown in No. 2. 1463. As long as the intervals between the piers are filled in with a wall, if that be placed flush with the outside, the piers will form pilasters inwards (see fig. 585.), as ifief, whose projection ef is equal to one half of the face he; this wall ought to have a thickness equal to he ; but if it is brought to the inner line of the face of the piers, they need be only two thirds of the thickness; so that the piers will form counterforts on the exterior. In conclusion, knowing the effort of the thrust, the calculations will not be attended with difficulty in providing against it by adequate means of resistance.

ON THE MODEL of A CoVED WAULT.

1464. The model (fig. 587. Nos. 1. and 2.) is a square on the plan, each of whose sides is 9 inches internal measure, enclosed by a wall 10 inches high to the springing of the vault, which is semicircular in form, and the voussoirs throughout 9 inches thick. It is divided into seven parts at the points of greatest effort, as shown in Nos. 1. and 2 in the plan and section. On one of the sides of the first is supposed to be traced the mean circumference TKG, the tangents FT, FG, the secant FO, the horizontal line IKL, and the verticals Bi and MK. We may now therefore consider this vault as four triangular pieces of cylindrical arches, each resting throughout the length of their base on one of the walls which forms the sides of the square. As the portions of arches or vaults are equal, it is only necessary to take one of them for an example.

1465. In the last example, cubes are taken instead of the surfaces, and surfaces instead of lines. Thus expressing the length of the wall by f, its height by a, and its thickness by a ; the arm of the lever

[merged small][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors]
[graphic]
[graphic]
[ocr errors]

2p || 2pd-2nc. 55 b - V't"::::=#.

1467. The horizontal effort of the upper part, represented by the line KL, will be expressed by the triangle eBa of the plan; that of the lower part iK in the section will be expressed by the trapezium eBCd on the plan.

1468. The plan of the vault being square, the base ed will be double EG = KL of the section; and the area of the triangle eEd equal to the square of KL=41 x 41j = 17104.

1469. Ea of the plan being equal to the square of 54 less the square of 41 #, that is, # the superior effort being 1710, their difference is 504, which being multiplied by the thickness of the vault, or 9, is 4536 for the expression of the thrust represented by p in the formula, and for that of

[ocr errors]

d, which represents TI, being 411, 2pd=37.51.92. 1470. To obtain the vertical effort of the upper part of the arch represented by m, its cube must be multiplied by KM, and the product divided by the arc KG. 1471. The cube of this part is equal to the curved surface passing through the middle of its thickness multiplied by the thickness. The mean area is equal to the product of the length nq taken on the plan multiplied by KM. mq being 117, and KM 17), the product expressing such mean area is 2005;, which multiplied by 9 makes the cube 18051} . This cube again multiplied by KM= 17}, and divided by the length of arc KG=46, gives 6727 for the value of m, and for 2m 13454; e being 12'. 2mc=1701005.

[ocr errors]

af 120 x 108

b. representing the vertical effort of the half vault, will be expressed by the cube multiplied by Bf=58), and divided by the mean circumference TKG =92.

1472. To obtain the cube, the mean superficies, that is, nq x Bf or 117 x 58}, is to be multiplied by the thickness AB =9, which gives 6844 x 9=61600).

This cube multiplied by Bf=584 and divided by the mean circu's TKG =92, that is, 61600 x # =3916988, for the value of b, and for that of a' #=302 and #=9 | 2. Substituting these values in the formula,

[ocr errors][ocr errors]

that is, a little less than 73 lines for the thickness of the walls, which is less than that of the vault; and shows that by giving the walls the same thickness as the vault, all the requisite solidity will be obtained. This is proved by experiments, for in the model the vault was borne equally well on walls of 9 lines in thickness divided into 8 parts, as upon 12 Doric columns whose diameter was 9 lines, four being placed at the angles and eight others under the lower part of the vault.

1473. To find the thickness of these walls by the geometrical method: Take the difference between the area of the triangle BEC and that of the triangle Eed, which divide by the length BC. 108 x 54

x

Thus, the area of the greater triangle being "#"=2916; that of the smaller one, #"*-17104, their difference, 12056 divided by 108–1116, which transfer to the

profile from B to h, and make Bn equal to the thickness of the vault. Upon nh, as a diameter, describe a semicircle, which at its intersection with the horizontal line BE will determine the thickness of the vault, and be found to be 10 lines. 1474. The small thrust of this species of vaulting occurs on account of the upper part, which causes it, diminishing in volume in proportion as the horizontal effort becomes more considerable, and because the triangular form of its parts and their position give it the advantage of having the larger sides for bases; whilst in groined vaulting, the triangular parts resting only on an angle, the weight increases as the horizontal efforts. 1475. Moreover, as the return sides mutually sustain each other, a half vault, or even a quarter vault, on a square base, would ": if the walls were 10 lines thick, proving that e 3

the opposite parts, acting little more than against each other, the thrust becomes almost nothing.

: By the method of the centres of gravity, Rondelet found a result less than that above given; but that arose from neglecting some points in the calculation which it was difficult to introduce for general practice.

1477. It is obvious that in the above application great allowance must be made when the apartment to be vaulted is not square; that is, its advantages diminish *15 the two opposite sides become longer than the width, and when the length is twice the width, or even much less, the thrusts must be calculated on the principle of cylindrical vaulting; and as in this species of vaulting the greatest effort occurs in or towards the middle of the sides, opening for doors and windows should there be avoided.

Application of the Method to Spherical Vaulting. 1478. The models (fig. 588. Nos. 1. and 2. and fig. 589. Nos. 1. and 2.) were of the

Fig. 588. Fig. 589.

same opening as the last mentioned. They are cut into eight equal parts by vertical planes crossing each other in the axis; each of these parts is subdivided by a joint at 45 degrees, altogether forming sixteen pieces. The vault stands on a circular wall of the same thickness divided into eight parts corresponding to those of the vault. All the parts are so arranged as to form continued joints without any bond, in order to give the experiment the most disadvantageous result. Yet it stood firmly, and was even capable of bearing a weight on the top. 1479. If for these eight pieces of circular wall we substitute eight columns of equal height, as in No. 1, fig. 589, so that the vertical joints fall over the middle of each column; the vault will still stand, although the cube of these columns, as well as their weight, occupies only one ninth part of the circular wall for which they are substituted. From this it is evident that spherical vaults have less thrusts than coved vaults. 1480. Applying the method of the preceding examples, describe the mean circumference (fig. 588. Nos. 1. and 2.), draw the tangents TF, GF, the secant FO, the horizontal line IKL, and the verticals KM and Bi; lastly, calculating for one eighth of the vault, take the sector Ohm to express the horizontal effort indicated by KL, and the part Hh Mn to express the horizontal effort of the lower part. 1481. The difference of these areas multiplied by the thickness of the vault will be the expression of the thrust p of the formula. 1482. The radius Om of the sector being 41i, and its length 32, its area will be 672: 1483. The area of hPIMm will be equal to the difference of the two sectors OHM and Ohm, whereof the first is equal to the product of half OM=27 by the arc HM=42, or 11454, the second =6723, , whence the difference =473:. 1484. The thrust, being equal to the difference between 672; and 473:, will be 198:49; therefore p =78633

[graphic]
[graphic]
« ZurückWeiter »