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 supported at both ends; but with the side upwards, when it is supported only at one end, (art. 252 vol. 2), 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 Cylinder 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 Cylin der which, when supported in like manner at one end, shall just bear without breaking another given weight at the unsupported end. L3 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 3:13:: ww, 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 lu 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 w or w, will be (w+U)L 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 vol. 2), or as 3 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 (‡w + u)l: (—/w +U)L :: 13: L3; this analogy, turned into an equation, gives L3 w+24 112 + 213ʊ=0, a cubic equa L3 213 พ 2 tion from which the numeral value of L may be easily determined, when those of the other letters are known. w+2u -IL2, or L = w + 2u Corol. 1. When u vanishes, the equation gives L3 = 1, whence w: w +2u: : : L, for the length of the beam, which will but just support its own weight. w 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 PROBLEM 36. Given the Length and Weight of a Cylinder or Prism, fixed Horizontally as in the foregoing problem, and a weight which, when hung at a given point, Breaks the Prism: it is required to determine how much longer the Prism, of equal Diameter or of equal Breadth and Depth, may be extended before it Break, either by its own weight, or by the addition of any other adventitious weight. Let / denote the length of the given prism, w its weight, and u a weight attached to it at the distance d from the fixed end; also let L denote the required length of the other prism, and u the weight attached to it at the distance D. Now the strain occasioned by the weight of the first beam is wl, and that by the weight u at the distance d, is du, their sum + du being the whole strain. In like manner wL + DU is the strain on the second beam; but /: L:: W: =w the LW Τ l weight of this beam, theref. +DU its strain. But the WL2 strength of the beam, which is just sufficient to resist these strains, is the same in both cases; therefore WL2 + DU= wl + du, and hence, by reduction, the required length wl+2du-2DU). L = √(1 x w Corol. 1. When the lengthened beam just breaks by its own weight, then u = 0 or vanishes, and the required length becomes L = √(lx wl +2du). Corol. 2. Also when u vanishes, if d become = l, then + = 2√ is the required length. w+ Qu พ PROBLEM 37. Let AB be a beam moveable about the end A, so as to make any angle BAC with the plane of the horizon AC: it is required to determine the position of a prop or supporter DE of a given length, which shall sustain it with the greatest ease in any given position; also to ascertain the angle BAC when. the least force which can sustain AB, is greater than the least force in any other position. Let Let G be the centre oprette! beam; and draw Gn: perD. TO AK AG, ni to Gm, and AP to Dk. 16: r = AG, p = DL. = the weigh: of the beam 46, and anx. Then, by the mature of the parallelogran o. forces, en : em, or by sim. triangles, A6 : AN = 2 :: U : =, the force whict, acting at & in the direction mo, is suffición" to sustair the ho and, by the nature of the lever, AL : AG = 7 1. quisite force at G : the force capabic of in a direction perp. to AB or parallel to me ; aa AF: AE :, the force or pressure actua AE the given prop DE in a direction perp.to at force will manifestly be the least posset wet 4. upon DE is the greatest possible, whatever cut anga i at be, which is when the triangle ADI I Busca', side AD = AE, by an obvious corol. from the antige frat wat prob. 6 pa. 171 of this volume. Secondly, for a solution to the setter pot ýtis porreicama |