= This necessarily happens, from the equality of the weights, and the similarity of the positions, and actions of the whole in both cases. Thus, from the equality of the corresponding weights, at the like angles, the ratios of the weights, ab, bd, dh, he, &c. in the lower figure, are the very same as those, ab, bD. DH, He, &c. in the upper figure; and from the equality of the constant horizontal forces CH, ch, and the similarity of the positions, the corresponding vertical lines, denoting the weights, are equal, namely, ab ab, bD bd, DH= dh, &c The same may be said of the oblique lines also, ca, cb, &c. which being parallel to the beams ab, bc, &c. will denote the tensions of these in the direction of their length, the same as the oblique thrusts or pushes in the upper figures. Thus, all the corresponding weights and actions and positions, in the two situations, being exactly equal and similar, changing only drawing and tension for pushing and thrusting, the balance and equilibrium of the upper figure is still preserved the same in the hanging festoon or lower one.. Scholium. The same figure, it is evident, will also arise, if the same weights, i, k, l, m, n, be suspended at like distances, ab, bc, &c. on a thread, or cord, or chain, &c. having in itself little or no weight. For the equality of the weights, and their directions and distances, will put the whole line, when they come to equilibrium, into the same festoon shape of figure. So that, whatever properties are inferred in the corollaries to the foregoing prob. will equally apply to the festoon or lower figure hanging in equilibrio. This is a most useful principle in all cases of equilibriums, especially to the mere practical mechanist, and enables him in an experimental way to resolve problems, which the best mathematicians have found it no easy matter to effect by VOL. II. Unu mere mere computation. For thus, in a simple and easy way he obtains the shape of an equilibrated arch or bridge; and thus also he readily obtains the positions of the rafters in the frame of an equilibrated curb or mansard roof; a single instance of which may serve to show the extent and uses to which it may be applied. Thus, if it should be required to make a curb frame roof having a given width AE, and consisting of four rafters AB, BC, CD, DE, which shall either be equal or in any given proportion to each other. There can be no doubt but that the best form of the roof will be that which puts B all its parts in equilibrio, so that there may be no unbalanced parts which may require the aid of ties or stays to keep the frame in its position. Here the mechanic has nothing to do but to take four like but small pieces, that are either equal or in the same given proportions as those proposed, and connect them closely together at the joints A, B, C, D, E. by pins or strings, so as to be freely moveable about them; then suspend this from two pins a, e, fixed in a horizontal line, and the chain of the pieces will arrange itself in such festoon or form, abcde, that all its parts will come to rest in equilibrio. Then, by inverting the figure, it will exhibit the form and frame of a curb roof abyde, which will also be in equilibrio, the thrusts of the pieces now balancing each b other, in the same manner as was done by the mutual pulls or tensions of the hanging festoon a bed e. By varying the distance ae, of the points of suspension, moving them nearer to, or father off, the chain will take different forms; then the frame ABCDE may be made similar to that form which has the most pleasing or convenient shape, found above as a model. Indeed this principle is exceeding fruitful in its practical consequences. It is easy to perceive that it contains the whole theory of the construction of arches: for each stone of an arch may be considered as one of the rafters or beams in the foregoing frames, since the whole is sustained by the mere principle of equilibration, and the method, in its application, will afford some elegant and simple solutions of the most difficult cases of this important problem. PROBLEM PROBLEM 32. Of all Hollow Cylinders, whose Lengths and the Diame" ters of the Inner and Outer Circles continue the same, it is required to show what will be the Position of the Inner Circle when the Cylinder is the Strongest Laterally. Since the magnitude of the two circles are constant, the area of the solid space included between their two circumferences, will be the same, whatever be the position of the inner circle, that is, there is the same number of fibres to be broken, and in this respect the strength will be always the -same. The strength then can only vary according to the situation of the centre of gravity of the solid part, and this again will depend on the place where the cylinder must first break, or on the manner in which it is fixed. Now, by cor 8 art. 251 Sta tics, the cylinder is strongest when the hollow, or inner circle, is nearest to that side where the fracture is to end, that is, at the bottom when it breaks first at the upper side, or when the cylinder is fixed only at one end as in the first figure. But the reverse will be the case when the cylinder is fixed at both ends; and con sequently when it opens first below, or ends above, as in the 2d figure annexed. PROBLEM 33. To determine the Dimensions of the Strongest Rectangular Beam, that can be cut out of a Given Cylinder. Let AB, the breadth of the required AC2— AB2, or d2—D2 —b2 ; D E F B imum ; in fluxions D3¿ — 362¿ =0 = ̧2—362, or d2 = 262; also d2 D2 —b2: = = 36a — b* = 262. Conseq. b2 : d2 : D2 :: 12: 3, that is, the squares of the breadth, and of the depth, and of the cylinder's diameter, are to one another respectively as the three numbers 1, 2, 3. Corol. Corol. 1. Hence results this easy practical construction: divide the diameter AC into three equal parts, at the points E, F ; erect the perpendiculars EB FD; and join the points B, D to the extremities of the diameter: so shall ABCD be the rectangular end of the beam as required. For, because AE, AB, AC are in continued proportion (theor. 87 Geom.), the ref. AE AC: in like manner AF: AC AC2 :: 1: 2: 3. AB: AD E AB2: AC2; and AD. AC; hence AE AF: AC :: Corol. 2. The ratios of the three b, d, D, being as the three ✔ 1, √2, 8, or as I, 1.414 1732, are nearly as the three 5, 7, 86, or more nearly as 12, 17, 20.8. Corol. 3. A square beam cut out of the same cylinder, would have its side = Dv=D √2. And its solidity would be to that of the strongest beam, as D2 to D2 2, or as 3 to 2/2, or as 3 to 2-828; while its strength would be to that of the strongest beam, as (D1)3 to D√ × D2, or as Į√2 to 3, or as 9√2 to 83, or nearly as 101 to 110. Corol. 4. Either of these beams will exert the greatest lateral strength, when the diagonal of its end is placed vertically, by art. 252 Statics. Corol. 5. The strength of the whole cylinder will be to that of the square beam, when placed with its diagonal vertically, as the area of the circle to that of its inscribed square. For, the centre of the circle will be the centre of gravity of both beams, and is at the distance of the radius from the lowest point in each of them; conseq. their strenths will be as their areas, by art. 243 Statics. PROBLEM 34. To determine the Difference in the Strength of a Triangular Beam, according as it lies with the Edge or with the Flat Side Upwards. In the same beam, the area is the same, and therefore the strength can only vary with the distance of the centre of gravity from the highest or lowest point; but, in a triangle,the distance of the centre of gravity from an angle, is double of its distance from the opposite side; therefore the strength of the beam will be as 2 to 1 with the different sides upwards, under different circumstances, viz. when the centre of gravity is farthest from the place where fracture ends, by art. 243 Statics, that is, with the angle upwards when the beam is ป supported supported at both ends; but with the side upwards, when it is supported only at one end, (art. 252 Statics), because in the former case the beam breaks first below, but the reverse in the latter case. PROBLEM 35. Given the Length and Weight of a Clylinder or Prism, placed Horizontally with one end firmly fixed, and will just support a given weight at the other end without breaking; it is required to find the Length of a Similar Prism or Cylinder which, when supported in like manner at one end, shall just bear without breaking another given weight at the unsupported end Let denote the length of the given cylinder or prism, d the diameter or depth of its end, w its weight, and u the weight hanging at the unsupported end; also let the like capitals L, D, w, u denote the corresponding particulars of the other prism or cylinder. Then, the weights of similar solids of the same matter being as the cubes of their lengths, as 13:13:: w3w, the weight of the prism whose length is L. Now wl will be the stress on the first beam by its own weight w acting at its centre of gravity, or at half its length; and in the stress of the added weight u at its extremity, their sum (w+u) will therefore be the whole stress on the given beam in like manner the whole stress on the other beam, whose weight is worw, will be (w+U) or (w+U)L. L3 L3 But the lateral strength of the first beam is to that of the second, as d3 to D3 (art. 246 Statics), or as 13 to L3; and the strengths and stresses of the two beams must be in the same ratio, to answer the conditions of the problem; therefore as L3 2/3 (\w+u)¡ : (773w+U)L :: 13: L3; this analogy, turned into w+2u, 2 an equation, gives 13- -/L2+= 13 U = 0, a cubic equa w tion from which the numeral value of L may be easily determined, when those of the other letters are known. Corol. 1. When u vanishes, the equation gives L3 = w+zu, พ w+2u "/L2, or L == ", whence w: w+2u ::/: L, for the length of the beam, which will but just support its own weight. Corol. 2. If a beam just only support its own weight, when fixed at one end; then a beam of double its length, fixed at both ends, will also just sustain itself: or if the one just break, the other will do the same. PROBLEM |