inches deep, can support a weight of 2000 libs; another beam of the same material, 6 feet long, 4 inches broad, and 4 inches deep, will support double that weight, being directly as their breadths;-but a beam of that material, 6 feet long, 2 inches broad, and 8 inches deep, will sustain a weight of 8000 libs.; being as the square of their depths. From a mean of experiments made, to ascertain the transverse strength of various bodies, it appears that the ultimate strength of an inch square, and an inch round bar of each, 1 foot long, loaded in the middle, and laying loose at both ends, is nearly as follows, in libs. avoirdupois. Names of Bodies. Sq. Bar. 1 third. Rd. Bar. 1 third. To find the ultimate transverse strength of any rectangular beam, supported at both ends, and loaded in the middle; or supported in the middle, and loaded at both ends; also, when the weight is between the middle and the end; likewise, when fixed at one end and loaded at the other. RULE.-Multiply the strength of an inch square bar, 1 foot long, (as in the table,) by the breadth, and square of the depth in inches, and divide the product by the length in feet; the quotient will be the weight in libs. avoirdupois nearly. EXAMPLE 1.-What weight will break a beam of oak 4 inches broad, 8 inches deep, and 20 feet between the supports? When a beam is supported in the middle and loaded at each end, it will bear the same weight as when supported at both ends and loaded in the middle, that is, each end will bear half the weight. When the weight is not situated in the middle of the beam, but placed somewhere between the middle and the end,-Multiply twice the length of the long end, by twice the length of the short end, and divide the product by the whole length of the beam; the quotient will be the effectual length. EXAMPLE 2-Required the ultimate transverse strength of a pitch pine plank, 24 feet long, 3 inches broad, 7 inches deep; and the weight placed 8 feet from one end. Again, when a beam is fixed at one end and loaded at the other, it will only bear of the weight as when supported at both ends and loaded in the middle. EXAMPLE 3. What is the weight requisite to break a deal beam 6 inches broad, 9 inches deep, and projecting 12 feet from the wall? 566 × 6 × 92 nearly. 12 2292345730.7 libs. The same rules apply as well to beams of a cylindrical form, with this exception, that the strength of a round bar (as in the table) is multiplied by the cube of the diameter, in place of the breadth, and square of the depth. EXAMPLE 4.-Required the ultimate transverse strength of a solid cylinder of cast iron, 12 feet long, and 5 inches diameter. EXAMPLE 5.-What is the ultimate transverse strength of a hollow shaft of cast iron, 12 feet long, 8 inches diameter, outside, and containing the same cross sectional area as a solid cylinder, 5 inches diameter ? 82 526.24. and 83 6.243 = 269. then, 2026 × 269 12 = 45416 libs. nearly. NOTE.-When a beam is fixed at both ends and loaded in the middle, it will bear one-half more than it will when loose at both ends. And if a beam is loose at both ends, and the weight laid uniformly along its length, it will bear double; but, if fixed at both ends, and the weight laid uniformly along its length, it will bear triple the weight. PROBLEM II. To find the breadth or depth of beams intended to support a permanent weight. RULE.-Multiply the length between the supports, in feet, by the weight to be supported in libs, and divide the product by one-third of the ultimate strength of an inch bar, (as in the table,) multiplied by the square of the depth; the quotient will be the breadth, or, multiplied by the breadth, the quotient will be the square of the depth; both in inches. EXAMPLE 1.-Required the breadth of a cast iron beam, 16 feet long, 7 inches deep, and to support a weight of 4 tons in the middle. 4 tons 8960 libs, and 8960 × 16 860 × 72 = 3.4 inches nearly. EXAMPLE 2.-What must be the depth of a cast iron beam 3.4 inches broad, 16 feet long, and to bear a permanent weight of 4 tons in the middle ? NOTE 1.-When a beam is fixed at both ends, the divisor must be multiplied by 1.5 on account of it being capable of bearing one-half more. 2. When a beam is loaded uniformly throughout, and loose at both ends, the divisor must be multiplied by 2, because it will bear double the weight. 3. If a beam is fast at both ends, and loaded uniformly throughout, the divisor must be multiplied by 3, on account that it will bear triple the weight. EXAMPLE 3.-Required the breadth of an oak beam, 20 feet long, 12 inches deep, made fast at both ends, and to be capable of supporting a weight of 12 tons in the middle. 12 tons 26880 libs. and 26880 × 20 266 x 122 x 1.5 = 9.7 inches nearly. Again, when a beam is fixed at one end, and loaded at the other, the divisor must be multiplied by .25; because it will only bear one-fourth of the weight. EXAMPLE 4.-Required the depth of a beam of ash, 6 inches broad, 9 feet projecting from the wall, and to carry a weight of 47 cwt. 5264 × 9 379 x 6 x .25 = 9.12 inches deep nearly. And when the weight is not placed in the middle of a beam, the effective length must be found as in problem first. EXAMPLE 5.-Required the depth of a deal beam 20 feet long, and to support a weight of 63 cwt. 6 feet from one end. 188 × 6 16.8 effective length of beam, and 7056 libs. hence 10.24 inches deep nearly. Beams or shafts exposed to lateral pressure are subject to all the foregoing rules, but in the case of water-wheel shafts, &c., some allowance must be made for wear, then the divisor may be changed from 675 to 600 for cast iron. EXAMPLE 6.-Required the diameter of bearings for a water-wheel shaft, 12 feet long, to carry a weight of 10 tons in the middle. And when the weight is equally distributed along its length, the cube root of half the quotient will be the diameter, thus, 448 2 = 3√√√/224 = 6.07 inches diameter nearly. solid EXAMPLE 7.-Required the diameter of a cylinder of cast iron, for the shaft of a crane, to be capable of sustaining a weight of 10 tons; one end of the shaft to be made fast in the ground, the other to project 6 feet; and the effective leverage of the jib as 13 to 1. 10 tons =22400 libs., and 22400 × 6.5 x 1.75 3 675 ×.25 =1509 And 1509 11.47 inches diameter nearly. The strength of cast iron to wrought iron, in this direction, is as 9 is to 14 nearly; hence, if wrought |