منو 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 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 a 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 A E other, in the same manner as was done by the mutual pulls or tensions of the hanging festoon abcde. By varying the distance ae, of the points of suspension, moving them nearer to, or farther 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 v.2, 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. F Let AB, the breadth of the required beam, be denoted by b, AD the depth by d, and the diameter AC of the cylinder by D. Now when AB is horizontal, the lateral strength is denoted by bd (by art. 248 vol. 2), which is to be a maximum. But AD AC2. AB2, or d2 = D - b2; theref. bd2(D2-b2)b=d3b-b3 is a maximum: in fluxions Db36b0 D2 - 352, or D2 = 362; also d2 D2 62 = 362 b2b. Conseq. b2 : d2 : D'::' 23, 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. Z-2 = -- Corcl B C mere computation. For thus, in a simple and easy way he - obtains the shape of an equilibrated arch or bridge; and thus also be 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 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 a 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 alyde, which will also be in equilibrio, the thrusts of the pieces now balancing each A E other, in the same manner as was done by the mutual pulls or tensions of the hanging festoon abcde. By varying the distance ae, of the points of suspension, moving them nearer to, or farther 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. 330 Of all Hollore Cylinders, whose Lengths and the Diame ters of the Inner and Outer Circles continue the ame required to shore what will be the Position of the Inner Circle when the Cylinder is the Strongest. Later iłły Since the magnitude of the wo troles are cunostint area of the solid wace, incluted setween hete ferences, will be the same, whaterem je inner circle, that is, here is he ame lumber broken, and in mis remet he frenoth wil same. The strength then an inly men wounding situation of the centre of gravity of he old again will tenend in he place where he mullacer met break, or in the manner n auch reixest is Teareer a inder 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 pro F portion (theor. 87 Geom.), theref. AE: AC :: AB2: AC2; and in like manner AF: AC:: AD2: Ac2; hence AE AF AC :: AB2: AD2: AC2 ;; 1: 2: 3. Corol. 2. The ratios of the three b, d, D, being as the three 1, 2, 3, or as 1, 1'414, 1·732, 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 =D/D√2. And its solidity would be to that of the strongest beam, as D2 to D2/2, or as 3 to 22, or as 3 to 2.828; while its strength would be to that of the strongest beam, as (D)33 to DX D2, or as √2 to 33, or as 9/2 to 8/3, 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 vol. 2. 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 strengths will be as their areas, by art. 243 vol. 2. 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 vol. 2; that is, with the angle upwards when the beam is supported |