2. Multiply also the cube of the length, in inches, by the given weight in lbs. ; then divide the latter product by the former for the deflection sought. EXAM. 1. A square beam of English oak, whose side is 6 inches, is supported on two walls, 20 feet distant, and is to be loaded at its middle point with 1000lbs., what will it be deflected? Ans. 1.8 inch. EXAM. 2. A beam of red pine, 8 inches in breadth, and 1 foot deep, is supported on two walls, distant 33 feet 4 inches: how much will it be deflected with 2000lbs. suspended at its centre? Ans. 14 inches. Note. If the beam be fixed at each end, the deflexion will, with equal weights, be two-thirds of that found by the above rule. PROB. IV. To compute the Deflection of Beams supported at each end, and loaded uniformly throughout their Length with a given Weight. Rule. Compute the deflection the same as in the last problem. Multiply that result by 5, and divide the product by 8, and the quotient will be the answer. EXAM. 1. A uniform bar of Adriatic oak, 2 inches square, is rested upon two props, distant 24 feet, how much will it be deflected by its own weight, its specific gravity being 960, or 60lbs. to the cubic foot? Ans. 91⁄2 inches. EXAM. 2. A beam of Riga fir, 12 inches square, is to support the brick work over a gateway, 12 feet wide; the computed weight of the brick work is 30000lbs., what deflection may be expected? Ans. 58 inch. PROB. V. To compute the ultmate Deflection of Beams, or Rods, before their Rupture. Note. The beams are supposed to be supported at each end. Rule. Multiply the tabular value of u, in the preceding table of data, by the depth of the beam in inches, and divide the square of the length, also in inches, by that product, for the ultimate deflection sought. EXAM. A square inch rod of ash, 6 feet long, is broken by a weight applied to its centre: how much will it be deflected before it breaks? Ans. 13.1 inches. PROB. VI. To find the ultimate transverse Strength of any rectangular Beam of Timber, fixed at one End and loaded at the other. Rule 1. Multiply the value of s, in the preceding table of data, by the breadth and square of the depth, both in inches, and divide that product by the length, also in inches, and the quotient will be the weight in lbs. This is approximative. Rule. II. 1. Take the ultimate deflection 8 times that of the last problem, and divide the deflection by the length, which will give the sine of the angle of deflection; whence, by a table, find the secant. 2. Multiply this secant by the breadth and square of the depth in inches, and the product again by the value of s' in the table of data. 3. Divide this last product by the length in inches, and the quotient will be the answer, in lbs. EXAM. 1. What weight will it require to break a piece of Mar forest fir, fixed by one end in a wall, and loaded at the other; the breadth being 2 inches, depth 3 inches, and length 4 feet? Ans. 518lbs. EXAM. 2. A square oaken balk, 12 inches square, projects 8 feet 4 inches from a solid wall, in which it is fixed; what weight will be sufficient to break it ? Ans. 50345lbs. EXAM. 3. A piece of ash, 2 inches square, projects 6 feet from a wall in which it is fixed; what weight, uniformly distributed through its length, will be required to break it ? PROB. VII. To compute the ultimate transverse Strength of any rectangular Beam, when supported at both Ends and loaded in the Centre. Rule 1. Multiply the tabular value of s by 4 times the breadth and square of the depth in inches, and divide that product by the length, also in inches, for the weight. Rule 11. 1. Compute the ultimate deflection by Prob. v. ; square that deflection, and divide it by the square of half the fength of the beam, and add the quotient to 1, for the square of the secant of deflection; which multiply by the length in inches. 2. Multiply the tabular value of s' by 4 times the breadth, and the square of the depth; and divide that product by the former, for the answer in lbs. EXAM. What weight will be necessary to break a piece of larch similar to the 3rd specimen, the length being 8 feet 4 inches, the breadth 8 inches, and depth 10 inches; being supported at each end, and loaded in the middle ? Ans. 36676lbs. Note 1. When the beam is loaded uniformly throughout its length, the same rule will apply, but the result must be doubled. 2. If the beam be fixed at each end and loaded in the middle, then the result obtained in the problem must be increased by its half. 3. If the beam be fixed at both ends and loaded uniformly throughout its length, the same result must be multiplied by 3. That is, the strength under these several circumstances: 1 :2 are 2 :4 14:3 3 :6 Supported and loaded in the centre... as EXAM. A piece of New England fir, 10 feet long and 6 inches square, being fixed at each end, and loaded uniformly through its entire length: it is required to find the weight necessary to break it. Ans. 24036lbs. PROB. VIII. To find the Weight under which a Column of Timber of given Dimensions and Elasticity will begin to bend, when placed, vertically, on a horizontal Plane. Rule. Multiply into one sum the value of E for the proposed wood, the cube of the least thickness, and the greatest thickness, the two latter both in inches; and that product again by the constant number -2056. Then divide the last product by the square of the length, in inches, for the an. swer, or weight in lbs.* * This rule is founded upon the formulæ which have been given for this particular case, by Euler, Poisson, &c. 36 Ekk #pf 4f2 126 = the weight, under which a column begins to bend. Where is half the weight, f half the length, and b the deflection, when the beam or column is loaded in the middle, and supported at its two ends: also, = 3.14159, &c. or the semicircumference of a w(l)3 38 Circle to radius 1; that is, according to our notation, ekk = , or EXAM. 1. What weight will be requisite to bend a rod of red pine, 10 inches in length and 1 inch square, when placed vertically on a plane, the weight being applied at its upper extremity? Ans. 15131lbs. EXAM. 2. Assuming the elasticity of English oak at 5806200, what weight will it require to bend a column, 8 feet 4 inches in length and 10 inches square? Ans. 1193754lbs. EXAM. 3. What weight will it require to bend a column of the same wood, and the same lateral dimensions, but of double the length? Ans. 298438lbs. PRACTICAL QUESTIONS. QUESTION I. A LARGE vessel, of 10 feet, or any other given depth, and of any shape, being kept constantly full of water, by means of a supplying cock, at the top; it is proposed to assign the place where a small hole must be made in the side of it, so that the water may spout through it to the greatest distance on the plane of the base. Let AB denote the height or side of the vessel; D the required hole in the. side, from which the water spouts, in the parabolic curve DG, to the greatest distance BG, on the horizontal plane. A D G B By the scholium art. 268, Hydraulics, the distance BGisalways equal to 2(AD. DB), which is equal to 2[x(a-x)] or 2(ax-x2), if a be put to denote the whole height AB of the vessel, and x = AD the depth of the hole. Hence 2 (ax - x2), or ax 23, must be a maximum. In Auxions, aż - 2x = 0, or a 2x = 0, and 2x = a, or ing this in our second formula for Ekk, and I for 2f, we have 4813 נן which is the same as the rule in words. x = 1a. So that the hole D must be in the middle between the top and bottom; the same as before found at the end of the scholium above quoted. QUESTION II. If the same vessel as in QUEST. 1, stand on high, with its bottom a given height above a horizontal plane below; it is proposed to determine where the small hole must be made, so as to spout farthest on the said plane. Let the annexed figure represent the vessel as before, and be the greatest distance spouted by the fluid, DG, on the plane bo. Here, as before, b = 2√(AD. Db) =2√[x(c-x)] = 2√(cx - x2), by putting ab = c, and AD = x. So G A D B h And that 2(cx- x2) or cr - x2 must be a maximum. hence, like as in the former question, x==ab. So that the hole must be made in the middle between the top of the vessel, and the given plane, that the water may spout farthest. QUESTION III. But if the same vessel, as before, stand on the top of an inclined plane, making a given angle, as suppose of 30 degrees, with the horizon; it is proposed to determine the place of the small hole, so as the water may spout the farthest on the said inclined plane. Here again (D being the place of the hole, and BG the given inclined plane), bG=2(AD.Db) = 2√[(a−x+z)], putting z = Bb, and, as before, a = AB, and x = AD. Then be must still be a maximum, as also Bb, being in a given ratio to the maximum BG, on account of the given angle в. Therefore ar G A G D B دا 2 ± xz, as well as z, is a maximum. Hence, by art. 94 of the Fluxions, aż - 2xi ± zi = 0, or a - 2x + z = 0 ; conseq. ± z = 22-a; and hence bo = 2√x(a-x+z) becomes barely 27. But as the given angle Bob is = 30°, the sine of which is therefore BG = 2вь от 2z, and be3 = BG-Bb2 = 322 = 3(2x - a)2, or bg = ± (2x - a) 3. Putting, now, these two values of be equal to each other, gives the equation 2x = (2x-a) 3, from which is found |