1ft. 7 in. 2 lines, for the thickness sought. 32.0 24.0 1545. If the walls supporting the roof were stiffened by extra means, such as lower roofs at an intermediate height, as in churches with a nave and side aisles, we may make Be in the diagonal BD (fig. 605.) equal to one twelfth of the height above the springing of the side roofs, and ef a twenty-fourth part of that height below it, and draw through the point ƒ a line parallel to AB, which will determine the thickness Af' sought; or, which amounts to the same thing, add together the total height AB of the interior, and that of E B above the point of support, E, whereof take the twenty-fourth part, which will be equal to Be + ef. Second Example. Fig. 604. 1546. Fig. 605. is a section of St. Paolo fuorì le murà, near Rome, as it was in 1816 The interior height to the under side of the tie-beams is 93 ft. 10 in. (French), whereof ft. 2 in. is the exterior height above the roofs of the side aisles. These two dimensions together make 120 ft., whose twenty-fourth part is 5 ft., to which, on the diagonal BD make By equal; then from the point f letting fall a vertical line, the horizontal line Be will determine the thickness, which will be 3 ft., the width of the nave being 73 ft. 6 in. In figures, as follows: BD BD= √93 ft. 10 in. × 93 ft. 10 in. + 73 ft. 6 in. x 73 ft. 6 in. = √14207=119 ft. 2 in. 1547. For the thickness, eB, as before, BD: AD:: Bf: Af'; whence, Af'AD B 73x53 ft. 1 in., instead of 2 ft. 11 in. 9 lines, the actual thickness of the walls. 1549. The same calculation being applied to the walls of the nave of Santa Sabina 119 ft. 2 in. -- Ꭰ 2 (Rome), whose height of nave is 51 ft. 2 in, and width 42 ft. 2 in., with a height of 16 ft. of wall above the side aisles, gives 21 in. 4 lines, and they are actually a little less than 24 in 1549. In the church of Santa Maria Maggiore, the width is 52 ft. 7 in., and 56 ft. 6 in and 4 lines high, to the ceiling under the roof. The height of the wall above the side aisles is 19 ft. 8 in., and the calculation requires the thickness of the walls to be 264 in. instead of 28 in., their actual thickness. 1550. In the church of St. Lorenzo, at Florence, the internal width of the nave is 37 ft. 9 in., and the height 69 ft. to the wooden ceiling; from the side aisles the wall is 18 ft. high. The result of the calculation is 21 in., and the actual execution 21 in. and 6 lines. 1551. The church of Santo Spirito, in the same city, which has a wooden ceiling sus pended to the trusses of the roof, is 76 ft. high and 37 ft. 4 in. wide in the nave the walls rise 19 ft. above the side aisles. From an application of the rule the thickness should b. 21 in. 3 lines, and their thickness is 22 in. 1552. In the church of St. Philippo Neri, at Naples, the calculation requires a thicknes of 21 in., their actual thickness being 22 in. 1553. In the churches here cited, the external walls are much thicker; which was ne. cessary, from the lower roofs being applied as leantoes, and hence having a tendency, in case of defective framing of them, to thrust out the external walls. Thus, in the church of St. Paolo, the walls are 7 ft. thick, their height 40 ft.; 3 ft. 4 in. only being the thickness required by the rule. A resistance is thus given capable of assisting the walls of the aisles, which are raised on isolated columns, and one which they require. 1554. In the church of Santa Sabina, the exterior wall, which is 26 ft. high, is, as the rule indicates, 26 in. thick; but the nave is flanked with a single aisle only on each side, and the walls of the nave are thicker in proportion to the height, and are not so high. For at St. Paolo the thickness of the walls is only of the interior width, whilst at Santa Sabina it is At San Lorenzo and San Spirito the introduction of the side chapels affords great assistance to the external walls. SECOND RULE. For the Thickness of Walls of Houses of more than one Story. 1555. As in the preceding case, the rules which Rondelet gives are the result of ob servations on a vast number of buildings that have been executed, so that the method proposed is founded on practice as well as on theory. 1556. In ordinary houses, wherein the height of the floors rarely exceeds 12 to 15 ft. in order to apportion the proper thickness to the interior or partition walls, we must b guided by the widths of the spaces they separate, and the number of floors they have to carry With respect to the external walls, their thickness will depend on the depth and height o the building. Thus a single house, as the phrase is, that is, only one set of apartments in depth, requires thicker external walls than a double house, that is, more than one apartmen in depth, of the same sort and height; because the stability is in the inverse ratio of the width 1557. Let us take the first of the two cases (fig. 606.), whose depth is 24 ft. and heigh pa to the under side of the roof 36 ft. Add to 24 ft. the half of the height, 18, and take of the sum 42, that is, 21 in., for the least thickness of each of the external walls above th set-off on the ground floor. For a mean stability add an inch, and for one still more soli add two inches. 1558. In the case of a double house (fig. 607.) with a depth of 42 ft., and of the san height as the preceding example, add half the height to the width of the building; that 21 to 18, and of the sum =194 is the thickness of the walls. To determine the thickne of the partition walls, add to their distance from each other the height of the story, a take of the sum. Thus, to find the thickness of the wall IK, which divides the spa LM into two parts and is 32 ft., add the height of the story, which we will take at 10 f making in all 42 ft., and take or 14 in. Half an inch may be added for each story abo he ground floor. Thus, where three stories occur above the ground floor, the thickness the lower one would be 15 in., a thickness which is well calculated for bricks and stone, whose hardness is of a mean description. 1559. For the wall AB, which divides the space between the external walls, equal to 35 ft., add to it the height, which is 10 ft., and of 45, the sum of the two; that is, 15 in. is the thickness required for the wall, if only to be carried up a single story; but if through more, then add half an inch, as before, for each story above the ground floor. For the spaces NO, PQ, RS, in this and the preceding figure, the repetition of the operation will give their thick nesses. 1560. To illustrate what has been said, fig. 608. is introduced to the reader, being Fig. 607. M the plan of a house in the Rue d'Enfer, near the Luxembourg, known as the Hotel Vendôme, built by Le Blond. It is given by D'Aviler in his Cours d'Architecture. The building is 46 ft. deep on the right side and 47 ft. in the middle, and is 33 ft. high from the pavement to the entablature. Hence, to obtain the thickness of the walls on the line FF, take the 47+33 sum of the height and width="2=40 ft., whose twenty-fourth part is 20 in. The building being one of solidity, let 2 in. be added, and we obtain 22 in. instead of 2 ft., which is their actual thickness. For the thickness of the interior wall, which crosses the building in the direction of its length, the space between the exterior walls being 42 ft. and the height of each story 14 ft., the thickness of this wall should be = 18 in. 8 lines, instead of 18 in., which the architect assigned to it. 42+14 1561. By the same mode of operation, we shall find that the thickness of the wall R, separating the salon, which is 22 ft. wide, from the dining-room, which is 18 ft. wide and 14 ft. high, should be 18 in. and 6 lines instead of 18 inches; but as the exterior walls, which are of wrought stone, are 2 ft. thick, and their stability greater than the rule requires, the interior will be found to have the requisite stability without any addition to their thickness. 1562. We shall conclude the observations under this head, by reference to a house built by Palladio for the brothers Mocenigo, of Venice, to be found in his works, and here given (fig. 609.). Most of the buildings of this master are vaulted below; but the one in question is not in that predicament. The width and height of the principal rooms is 16 ft., and they are separated by others only 8 ft. wide, so that the width which each wall separates is 254 ft., and their thickness consequently should be 25+ 16-13 in. 10 lines. The walls, as executed, 36 Fig. 609. hl nd are 14 in. in thickness. The exterio walls being 24 ft. high, and the depth c the building 46 ft. Their thickness b the rule should be 174 in.: they are 18 in On passing the Metropolitan Build ing Act in 1855, previous to whic the thicknesses of walls depended o buildings falling within certain classes o rates, we had the satisfaction of advising the Government to adopt the thicknesse of walls now directed to be used. Thes are based upon rules deduced from sec tions 1512 et seq. Inasmuch, however, a it was thought that builders might be liable to mistakes in extracting the squar root of the sum of the squares of th heights and lengths of walls, tables wer inserted in the Act to meet all cases. Generally the formula t = will be a useful guide in adjusting the thickness of walls in which t = thickness, h and I respectively the height and length, d the diagonal forme by the height and length, and n a constant determined by the nature of the building. I the tables for dwelling-houses, the constant multiplier (n) used was 22; for warehouses, 20 And but for the interference in committee of the present Right Hon. Member fo Oxfordshire (Mr. Henley), for what scientific reasons it is difficult to say, the constan multiplier for public buildings would have been 18. When his less than the constants are 27, 23, and 20 respectively. Of the Stability of Piers or Points of Support. N, M 1563. Let ABCD (fig. 610.) be a pier with a square base whose resistance is require in respect of a power at M acting to overturn it horizontally in the direction MA, or obliquely in that of NA upon the point D. Considering the solid reduced to a plane passing through G, the centre of gravity of the pier, and the point D, that upon which the power is supposed to cause it to turn, let fall from G the vertical cutting the base in I, to which we will suppose the weight of the pier suspended, and then supposing the pier removed, we only have to consider the angular lever BDI or HDI, whose arms are determined by perpendiculars drawn from the fulcrum D, in one direction vertical with the weight, and in the other perpendicular to the direction of the power acting upon the pier, according to the theory of the lever explained in a previous section. C D R Fig. 610. H RXID 1564. The direction of the weight R being always represented by a vertical let fall from the centre of gravity, the arm of its lever ID never changes, whatever the direction of the power and the height at which it is applied, whilst the arm of the lever of the power varie as its position and direction. That there may be equilibrium between the effort of th power and the resistance of the pier, in the first case, when the power M acts in an hori zontal direction, we have M: R::ID: DB, whence M x DB=Rx ID and M= If the direction of the power be oblique, as NA in the case of an equilibrium, N: R::II Rx ID. : DH; hence N x DH= Rx ID and N= 1565. Applying this in an example, let the height of the pier be 12 ft., its width 4 ft., an its thickness 1 ft. The weight R of the pier may be represented by its cube, and is ther fore 12 x 4 x 1 = 48. The arm of its lever ID will be 2, and we will take the horizont power M represented by DB at 12; with these values we shall have M: 48::2: 12; hend 48 × 2 M x 12-48 x 2 and M= =8. 12 DH That is, the effort of the horizontal power M should be equal to the weight of S cul feet of the materials whereof the pier is composed, to be in equilibrium. 1566. In respect of the oblique power which acts in the direction NA, supposing DI =73, we have N: 48::2: 71, whence N x 748 x 2, therefore N= 48 × 2 7 =13}, whilst tl expression of the horizontal power M was only 8 ft. ; but it must be observed, that the arm of the lever is 12, whilst that of the power N is but 7 ft.; but 13 x 7}=8 × 12=96, which is also equal to the resistance of the pier expressed by 12x4x2=96 It is more. over essential to observe, that, considering the power NA as the result of two others, MA and FA, the first acting horizontally from M against A, tends to overthrow the pier; whilst the second, acting vertically in the direction FÃ, partly modifies this effect by increasing the resistance of the pier. 1567. Suppose the power NA to make an angle of 53 degrees with the vertical AF, and of 37 degrees with the horizontal line AM; then NA: FA: MA::rad. : sin. 37 deg.: sin. 53 deg.::6: 10: 8. Hence, NA being found 134, we have 6: 10: 8::13:8:103. Whence it is evident that, from this resolution of the power NA, the resistance of the pier is increased by the effort of the power FA=8, which, acting on the point A in the direction FA, will make the arm of its lever CD=4, whence its effort = 8 x 4 = 32. 1568. The resistance of the pier, being thus found =96, becomes by the effort of the power FA=96+32=128. 1569. The effort of the horizontal power M being 103, and the arm of its lever being always 12, its effort 128 will be equal to the resistance of the pier, which proves that in this resolution we have, as before, the effort and the resistance equal. The application of this proposition is extremely useful in valuing exactly the effects of parts of buildings which become stable by means of oblique and lateral thrusts. 1570. If it be required to know what should be the increased width of the pier to counterpoise the vertical effort FA, its expression must be divided by ID, that is, 8 × 2, which gives 4 for this increased length, and for the expression of its resistance (12+ 4) × 4 × 2 =128, as above. 1571. If the effort of the power be known, and the thickness of a pier or wall whose height is known be sought so as to resist it, let the power and parts of the pier be represented by different letters, as follows. Calling the power p, the height of the pier d, the thickness sought ; if the power p act in an horizontal direction at the extremity of the wall or pier, its expression will be p x d. The resistance of the pier will be expressed by its area multiplied by its arm of lever, that is, d× ×; and supposing equilibrium, as the resistance must be equal to the thrust, we shall have the equation p × d=d × x × 1. Both sides of this equation being divisible by d, we have p= x; and as the second term is divided by 2, we obtain 2p=x x x or x2; that is, a square whose area =2p, and of which r is the side or root, or x= √2p, a formula which in all cases expresses the thickness to be given to the pier CD to resist a power M acting on its upper extremity in the horizontal direction MA. 1572. In this formula, the height of the pier need not be known to find the value of x, because this height, being common to the pier and the arm of the lever of the power, does not alter the result; for the cube of the pier, which represents its weight, increases or diminishes in the same ratio as the lever. Thus, if the height of the pier be 12, 15, or 24 ft., its thickness will nevertheless be the same. Example. — If the horizontal power expressed by p in the formula x= √2p be 8, we have = √16=4 for the thickness of the pier. Whilst the power acting at the extremity of the pier remains the same, the thickness is sufficient, whatever the height of the pier. Thus for a height of 12 ft. the effort of the power will be 8 × 12=96, and the resistance 12 x 4 x 2-96. If the pier be 15 ft. high, its resistance will be 15 × 4 × 2=120, and the effort of the power 8 x 15=120. Lastly, if the height be 24 ft., the resistance will be 24 × 4 × 2=192, and the effort of the power 8 x 24=192. I 1573. If the point on which the horizontal force acts is lower than the wall or pier, the difference may be represented by ƒ; and then px (d–f)=dx x x 2 ; Which becomes 2pd-2pf=dx2 and 2p-2 = x2; the formula becomes r=V18 18x6 = = √9=3, which is the thickness sought. 1574. When the power NA is oblique, the thickness may be equally well found by the arm of lever DH, by resolving it into two forces, as before. Thus, in the case of the oblique | power p=13}, calling ƒ its arm of lever 7, we shall have pxf=d, which will become f pf; in which, substituting the known values, we have z= whence x = √16 =4, the thickness sought of the pier. r, whence x = d 2 2x13x7 12 d |