« ZurückWeiter »
vAB31 AD2- v24 x 24+32 × 32; whence BD = V576 + 1024 = V1600 = 40 ft. Bb, which is the twelfth part of AD, or of 32 ft. = 2 ft. 8 in. ; the thickness of the wall expressed byAli", will be **- 13 ft., or 1 ft. 7 in. 2 lines, for the thickness sought. 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 equal to one twelfth of the height a above the springing of the side roofs, and efa Zoo - twenty-fourth part of that height below it, and Fig. 604. draw through the point f 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 EB above the point of support, E, whereof take the twenty-fourth part, which will be equal to Be + of
1546. Fig. 605. is a section of St. Paolo fuori le mura, 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 26 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 Bf 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 : —
=nor-3 ft. 1 in., instead of 2 ft. 11 in. 9 lines, the actual thickness of the walls.
1548. The same calculation being applied to the walls of the nave of Santa Sabina (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. 74 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 26, in. instead of 284 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 suspended 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 be 21 in. 3 lines, and their thickness is 22 in. 1552. In the church of St. Philippo Neri, at Naples, the calculation requires a thickness of 21 in., their actual thickness being 22 in. 1553. In the churches here cited, the external walls are much thicker; which was necessary, 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 or. At San Lorenzo and San Spiñto the introduction of the side chapels affords great assistance to the external walls.
Eramples for the Thickness of Walls of Houses of many Stories.
1555. As in the preceding case, the rules which Rondelet gives are the result of observations 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 be 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 of 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 apartment 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 height
to the under side of the roof 36 ft. Add to 24 ft. the half of the height, 18, and take or part of the sum 42, that is, 21 in., for the least thickness of each of the external walls above the set-off on the ground floor. For a mean stability add an inch, and for one still more solid add two inches. 1558. In the case of a double house (fig. 607.) with a depth of 42 ft., and of the same height as the preceding example, add half the height to the width of the building; that is, 21 to 18, and h of the sum = 19% is the thickness of the walls. To determine the thickness of the partition walls, add to their distance from each other the height of the story, and take 3, of the sum. Thus, to find the thickness of the wall IK, which divides the space LM into two parts and is 32 ft., add the height of the story, which we will take at 10 ft., making in all 42 ft., and take or 14 in. Half an inch may be added for each story above the ground floor. Thus, where three writyour above the ground floor, the thickness in
the lower one would be 15% in., a thick-
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 Daviller in his Cours d'Architecture. The building is
and their thickness consequently should be o: *=13 in 10 lines. The walls, as executed,
are actually 14 in. in thickness. The exterior walls being 24 ft. high, and the depth of the building 46 ft., their thickness by the rule should be 17 in. ; they are actually 18 in.
of the Stability of Piers, or Points of Support.
1563. Let ABCD (fig. 610.) be a pier with a square base whose resistance is required to be known in respect of a power at M acting upon it to overturn it horizontally in the direction MA, or obliquely in that of NA upon the point P. To render the demonstration more simple, we will consider 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. Letting 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 shall 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 Fig.610. acting upon the pier, according to the theory of the lever explained in a previous section.
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 varies as its position and direction. That there may be equilibrium between the effort of the 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 × DB = R x ID and M – “..."
... 1565. Applying this in an example, let the height of the pier be 12 ft., its width 4 ft., and its thickness 1 ft. The weight R of the pier may be represented by its cube, and is therefore 12 x 4 × 1 =48. The arm of its lever ID will be 2, and we will take the horizontal power M represented by DB at 12; with these values we shall have M : 48::2:12; hence M x 12 = 48 x 2 and M=***=s.
That is, the effort of the horizontal power M should be equal to the weight of 8 cube 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 DH
expression of the hozirontal 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 74 ft. ; but 13, x 74=8 x 12 = 96, which is also equal to the resistance of the pier expressed by 12 x 4 x 2 = 96. It is moreover 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 FA, 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. ::10:6 : 8. Hence, NA being found =1.84, we have 10 : 6 : 8:: 13; : 8 : 104. 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 EA, its expression must be divided by ID, that is, 8 x 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 a ; 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 x + x; ; and supposing equilibrium, as the p y 2 p g eq
resistance must be equal to the thrust, we shall have the equation p < d=dx r x . Both sides of this equation being divisible by d, we have p = r x: ; and as the second term is divided by 2, we obtain 2p =r x + or ao ; that is, a square whose area =2p, and of which a is the side or root, or r = 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 M.A. 1572. In this formula, the height of the pier need not be known to find the value of r, 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 r-wop be 8, we have a = v16=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 x 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 x 4 x 2 = 192, and the effort of the power 8 x 24 = 192. 1573. If the point on which the horizontal force acts is lower than the wall or pier, the difference may be represented by f: and then p < (d—f)=dx r x , ;