frame of a curb roof A B C D E, which will also be in equilibrio, the thrusts of the pieces now balancing each other, in the same manner as was done by the mutual pulls or tensions of the hanging festoon A b c d E.

4. If the mansard be constituted of four equal rafters; then, if angle C AE = m, angle c A B = x; it is demonstrable that 2 sin 2 x = sin 2 m. So that if the span ▲ E, and height м c, be given, it will be easy to compute the lengths ▲ B, B C, &c. Example. Suppose A E 24 feet, м c 12.

M C

Then = 1 = tan 45° angle c A M = m.

MA

... sin 2 m = sin 90° 1, and sin 2 x =

.. 2 x = 30°, and x = 15° = C A B Hence M A B = 45° + 15° = 60°

and MBA (180° — 2 × 15°)

=

and A M B= 180°

(75° +60°)

90° 15° = 75°

= 45°

Lastly, sin 75°: sin 45 :: A M = 12 : A B = 8·7846 feet.

Note. In this example, since A M = M C, as well as A B = BC, it is evident that м B bisects the right angle A м C; yet it seemed preferable to trace the steps of a general solution.

Stability of Arches.

But,

1. If the effect of the force of gravity upon the ponderating matter of an arch and pier, be considered apart from the operation of the cements which unite the stones, &c. the investigation is difficult to practical men, and it furnishes results that require much skill and care in their application. in an arch whose component parts are united with a very powerful cement, those parts do not give way in vertical columns, but by the separation of the entire mass, including arches and piers, into three, or, at most, into four parts. In this case, too, the conditions of equilibrium are easily expressed and easily applied.

Let f F, f' F', be the joints of rupture, or places at which the arch would most naturally separate, whether it yield in two pieces or in one.

Let G be the centre of gravity of the semi-arch ƒ F K k, and G' that of the pier A B F f. Let FI be drawn parallel to the horizon, and Gн be demitted perpendicularly

upon it; also let G' D be a perpendicular passing through G', and F E drawn from F parallel to it.

Then

2. PROP. If the arch fF F'f' tend to fall vertically in one piece, removing the sections ƒ F, f'r'; if a be the weight of the semi-arch fr K k, and P that of the pier up to the joint f F, the equilibrium will be determined by these two equations: viz.

where ƒ is the measure of the friction, or the tangent of the angle of repose of the material, and the first equation is that of the equilibrium of the horizontal thrusts, while the latter indi. cates the equilibrium of rotation about the exterior angle a of the pier.

3. PROP. If each of the two semi-arches F k, k, r', tend to turn about the vertex k of the arch, removing the points F, F', the equilibrium of horizontal translation, and of rotation, will be respectively determined by the following equations: viz.

4. Hence it will be easy to examine the stability of any cemented arch, upon the hypothesis of these two propositions. Assume different points, such as F in the arch, for which let the numerical values of the equations (1) and (2), or (3) and (4), be computed. To ensure stability, the first members of the respective equations must exceed the second; those parts will be weakest, where the excess is least.

If the figure be drawn on a smooth drawing pasteboard, upon a good sized scale, the places of the centres of gravity may be found experimentally, as well as the relative weights of the semi-arch and piers, and the measures of the several lines from the scale employed in the construction.

If the dimensions of the arch were given, and the thickness of the pier required, the same equations would serve, and different thicknesses of the pier might be assumed until the first members of the equations come out largest.

The same rules are applicable to domes, simply taking the ungulas instead of the profiles.

Models.

From an experiment made to ascertain the firmness of the model of a machine, or of an edifice, certain precautions are necessary before we can infer the firmness of the structure itself.

The classes of forces must be distinguished; as whether they tend to draw asunder the parts, to break them transversely, or crush them by compression. To the first class belongs the stretching suffered by key-stones, or bonds of vaults, &c. : to the second, the load which tends to bend or break horizontal or inclined beams; to the third the weight which presses vertically upon walls and columns.

PROB. 1. If the side of a model be to the corresponding side of the structure, as 1 to n, the stress which tends to draw asunder, or to break transversely the parts, increases from the smaller to the greater scale, as 1 to n3; while the resistance of those ruptures increases only as 1 to no.

The structure, therefore, will have so much less firmness than the model as n is greater.

If w be the greatest weight which one of the beams of the model can bear, and w the weight or stress which it actually sustains, then the limit of n will be n =

W

w

PROB. 2. The side of the model being to the corresponding side of the structure as 1 to n, the stress which tends to crush the parts by compression, increases from the smaller to the greater scale, as 1 to n3, while the resistance increases only in the ratio of 1 to n.

Hence, if w were the greatest load which a modular wall or column could carry, and w the weight with which it is actually loaded; then the greatest limit of increased dimensions would be found from the expression n = √

W

w

If, retaining the length or height n h, and the breadth n b, we wished to give to the solid such a thickness xt, as that it should not break in consequence of its increased dimensions, we should have x = n2

W

W

In the case of a pilaster with a square base, or of a cylin

drical column, if the dimension of the model were d, and of the largest pillar, which should not crash with its own weight when n times as high, x d, we should have

These theorems will often find their application in the profes sion of an architect or an engineer, whether civil or military.

3. Suppose, for an example, it were required to ascertain the strength of Mr. Smart's "Patent Mathematical Chain-bridge," from experiments made with a model. In this ingenious construction, the truss-work is carried across from pier to pier, so that the road-way from A to B, and thence entirely across, shall be in a horizontal plane, and all the base bars, diagonal bars, hanging bars, and connecting bolts, shall retain their own rective magnitudes throughout the structure. The annexed representation of half the bridge so exhibits the construction as to supersede the necessity of a minute verbal description.

Now, let represent the horizontal length of the model, (say 12 feet,) from interior to exterior of the two piers, w its weight (say 30 pounds), w the weight it will just sustain at its middle point в before it breaks (say 350 lbs.) Let n the length of a bridge actually constructed of the same material as the model, and all its dimensions similar: then, its weight will be n3 w, and its resisting power to that of the model, as n3 to 1, being n2 (w+w.) Hence n (w + 1⁄2 w) — § n3 w = no W - } n2 (n − 1) w, the load which the bridge itself would bear at the middle point.

=

Suppose n = 20, or the bridge 240 feet long, and entirely similar to the model; then we shall have (400 x 350) — 200 (20-1) 30 = 140000 114000 26000 lbs. 11 tons 124 cwt., the load it would just sustain in the middle point of its

extent.

--

=

=

Note. This bridge is, in fact, a suspension bridge, and would require brace or tie-chains at each pier. A considerable im provement upon its construction, by Colonel S. H. Long, of the American Engineers, is described in the Mechanic's Magazine, vol. xiii. or No. 368.