According a difference of something approaching 100 per cent.: Tredgold the highest, Clark the lowest, and Barlow about midway. New sets of experiments are desirable, especially for large scantlings. From 1866 D. Kirkaldy has made a number of experiments on fullsize timbers, and of late years Mr. Lanza has made some at the Massachusetts Institute. His results show that the ordinary formulæ require revision. For instance, a spruce beam 12 inches by 2 inches and 15 feet long broke with a central load of 5894 lbs. to Tredgold's formula, it ought to have carried a load of 8928 lbs. before breaking. This result was corroborated by other tests, and the general conclusion arrived at is that, whereas cb d2 we have been accustomed to use as a constant in the familiar formula W: 4 cwt. for fir or pine beams (in fact, in one of his examples 530 lbs. is used), we ought really to use a constant of not more than 2 cwt. The more thoroughly large size specimens, whether of wood or iron, are tested, the more will our knowledge of their strength be increased, and we shall be less dependent upon theories. (J. Slater, 1887). As an instance, put forward so long since as 1871-72 by Captain (now Colonel) Seddon from experiments by D. Kirkaldy on (1) white Riga fir, and (2) red Dantzic fir, where each piece was 20 feet longWhite Riga fir was 13 inches square=169 square inches area. Ultimate stress 331,260 lbs. 1960 lbs. per square inch. or 147 88 tons=126'04 tons per square foot. Deflection *642. Gave way at knots 2 feet 9 inches from centre. or 138:00 tons=112-02 tons per square foot. Gave way at knot 0-9 off centre. Deflection 548 only. Transactions, Royal Institute of British Architects, 1871-72, pp 156–164. Strictly speaking, the law that the breaking weight depended on the variations of stress was known before 1849, when it was shown that cast iron bars broke with half the statical breaking weight when subjected to continued repetitions of load. Subsequently Sir W. Fairbairn carried out an experiment on a riveted girder subjected to continual loading and unloading for a period of two or three years. It broke with two-fifths of the statical breaking load, and after repairing, with one-third the statical breaking weight, after over one million changes of load in each case. The statical breaking stress was not, as com monly assumed, the exact measure of the structural value of a material. in Builder, 1887, p. 741). (Prof. Unwin, 16299. PROBLEM I.-To find the breaking weight of a beam, the load being in the middle and all the dimensions known. The ends loose or supported. 359 lbs. 3.21 20 cwt. Here S, in the first formula, represents the value of the breaking weight in pounds in the middle, taken from the preceding table: in the other two it would be deduced from it; thus, taking Riga fir =3.21 for cwt., and 160 for tons: b breadth in inches; d depth in inches; distance between the points of support, in feet; a area of section. These letters will be continued in these problems, until other values are attached to them. W will always represent the breaking weight. Here F represents the weight in pounds borne by a rod 1 inch square, when the strain is as great as the rod will bear without destroying part of its elastic force, = 15,300 for cast iron. From a mean of 265 experiments by Hodgkinson and Fairbairn, it appears (by Gregory) that a weight of 454.4 pounds in the middle of a bar of cast-iron, 1 inch square and 4-5 feet bearing, produced fracture. Therefore, for a bar of any other dimensions we have : bd-F 61 × (1 − q × p3) = W lbs. Here q, difference between the breadth in the middle and the extreme breadth=625 as found to answer in practice; and p, depth of the narrow part in the middle 7 as found to answer. When the middle part of the beam is omitted, excep sufficient uprights to connect the top and bottom bars, then bd F 67 × (1-p3) = W lbs. Her Here 514 cwts. may be used for side castings, or 536 cwts, for erect castings. The other 514 cwts. 20 for tons 257, called 26 tons; 536 quantities are obtained thus: 26.8, called 27 tons. 1 When I is used in feet, 22=2166: a represents area of bottom flange in inches. Here a and d as before; P permanent load distributed, or about one-fourth of the breaking weight distributed; and multiplied by 2 when the ends are fixed = one-half BW. From the experiments above quoted from Gregory, we obtain— 16291. Gregory's work also states an arbitrary formula given by Mr. Dines, which he found to be tolerably correct in all cases where the length of the girder did not exceed 25 feet; its depth in the centre not greater than 20 inches; the breadth of the bottom flange not less than one-third, or more than half, the depth; the thickness of the metal not less thanth of the depth. Then 16 1792 [b d2 - (b-b2)d,2] = W lbs. | bd2-(b-b2)d.?] = W cwts. 0.8 [b [b d2-(b-b2)d2] = W tons. Here b entire breadth of bottom flange; b, thickness of the vertical part: d depth of whole girder; d, depth without the lower flange, all in inches; 7 length in feet. 24 1629m. Hurst, Handbook, notices that the area of the top flange should be of that of the bottom flange when the load is on the top; and when the load is on the bottom flange; Molesworth, Formula, has for the latter; he notes that if the depth of the girder be of the span, then a4 17 W tons, the weight being distributed. When the depth is 10 45= W tons, the weight being distributed. The depth at the ends may equal proximate rules for these girders have been given in the Pocket-book for 1865, as l x P = d × a, = a. Here I feet; P tons distributed; d depth of girder; a area of bottom flange, both in inches. ixa = P, d 1629n. For E, Wrought iron tube or beam, or box-beam :— 3 Ap Here a area of the bottom flange; C coefficient determined for this particular form of tube. In the table given by Fairbairn (pp. 116–17), a area of the whole cross section, the constant C17-8 tons, for a tube having the top flange = 142 thick, twice the area of the bottom one; the tube being 9.6 inches square, and 17.5 feet long between the supports. Such a beam deflected 1.76 inches with a breaking weight of 7,148 lbs. 12 16290. Hurst states it is usual to camber a riveted girder, so that on receiving the permanent load it may become nearly horizontal. If the required rise or camber equals e in the middle in inches, d being in inches and 7 in feet, we have K=e. For girders uniformly loaded and of uniform section throughout the length, K=018. When the section is also made to vary so that the girder will be of equal strength throughout, K·021. Molesworth notes the area of top flange as al 18; Hurst says al·75. If the depth of the girder be of the span, then W=133a tons; if, then W = 16a tons. The rivets to be inch and inch in diameter, placed 3 inches apart in the top and 4 inches apart in the bottom flange. 1629p. For F, Wrought-iron p'ate girder:— ad C-1500 W cwts.: area of the top flange being greater than the bottom flange, and the Winches adC 7 inches W tons, in which =W thickness of the web about or of the depth of the beam. ease C = 75 for deep plates, as 22 inches; cr=60 for less depth, as 7 inches. tons, when d is more than and the area of the top flange is 1.75 of the bottom flange. Here a area of bottom flange in inches; some calculators deduct the rivet holes from its total width. For depths under 12 inches, the width of the top flange should be half the depth; and when over 12 inches, one-third. With the latter proportion, feathers or stiffening pieces should be used to supply the deficiency in lateral stiffness occasioned by the reduced width of flange (par. 1629c.). The usual thickness of web for all depths under 3 feet is inch. Fairbairn (Tubular Bridges, p. 247) discovered that the top flange should have an area double that of the lower one to give the strongest form of wrought iron beam, a contrary principle to that obtained in cast iron. W x 8 WI 88d Wr W=150x1=100 16294. To find the area of either of the flanges at the centre of a girder supported at both ends, the formula is =a. Here W represents live and dead loads uniformly distributed in tons; I span in feet; d depth in inches; and s safe strain per square inch of metal on the flange, in tons. Therefore, say gxs-5x8:33 sect. area = 45 square inches, excluding rivet holes in the bottom flange. This formula is the equivalent of W weight per foot run x 12 α. When the sectional area of the top flange is to be greater 88 d than that of the bottom flange, multiply the latter area by 1.2. The average sectional area of a theoretically proportioned flanged girder may be taken at 3rds of the central sectional area. To find the sectional areas of either flange at any point along the whole length of the girder, the formula is (1−x) = a, excluding rivet holes in bottom flange. Here W weight per foot run in tons; a lesser segment into which the span is divided; 1- greater segment; s and d as before. To find the sectional area at any length of the web, the formula is = a square inches. Here x is the distance from the centre; Wand s as before. The vertical strain, at the centre of the beam, when one-half of the girder is fully loaded, is equal to of the fully loaded beam, that is W x 7. At the ends or pillars, the vertical strain is greatest, and is equal to W x 1. The strain at the centre, when the load is uniformly distributed, is obtained from the formula = 8. Here W distributed load in tons; I length in feet; d depth in feet; s strain in tons of compression in the top flange and tension in the bottom flange. Half the load collected at the centre of a girder being equal to the load distributed, the above formula becomes = 8. At any other point, ratios of strain will be as the square of half the span to the square of the segments into which a given point divides the span. The approximate strain at the centre, per square inch, on any beam, may be obtained from the formula =S. Here W distributed load in tons; I length in terms of the depth; a sectional area in square inches; and s strain in tons per square inch. To find the sectional area of a flange for a plate girder fixed at one end and free at the other the formula is =a, exclusive of rivet holes in the top flange; W weight in tons at the end of the girder; x length in feet from loaded end to the point where the sectional areas are required; d depth in feet; s safe strain per square inch in tons. When the load is uniformly distributed, using the same notations as before, except that W in tons is the load per foot run, the femula is =a. (Engineer's, Architect's, &c., Pocket-book,1865.) W x ds W x2 1629r. For G, Rolled irons or bars :— WI 8 a WI 4 d WI Here a area of bottom flange includes to above the upper part of the swelling for flange, as bc (fig. 6139.); or to the whole of the angle in plate girder F. A railway bar is often useful in country places; the intricate formula for the strength of the various parts will be found in Barlow; and in the Engineer's &c., Pocket-book for 1861. IIII Fig. 6134. 16298. For H, Tee irons, or Rolled T irons :— ad C=4 = W tons. 1629t. For I, Reversed Tee irons: —— When z is one-fourth of the table or flange bc, and the form as 5: 12 of a rectangle, then =3 N ـألـبـك SW. It was stated in the Oldham Mill Report that this fo m of beam, which might be considered to support a weight of say 1000 lbs., may be broken if reversed, that is, the flange placed uppermost, as T, with a weight of say 340 lbs. Hodgkinson experimented on two bars 4 feet 3 inches long, the flange 4 inches wide, rib 1 inch deep, with a thickness of metal of about inch. One bar was tried with the flange uppermost, the other bar with the flange downwards. The former broke with 2 cwt., the latter with 9 ewt. Experiments on three girders of this shape, the web being 2 inches high and inch thick. the flange 2 inches wide by inch thick, and 24 inches long, were made by Cooper of Drury Lane. He stated that the gain in strength over a flitch 2 inches by inch was 25 per cent.; the loss in stiffness being 30 per cent. The strength arising from the accumulation of the quantity submitted to tensile action bears out an adequate result, or 580 times its own weight, instead of 400, as 2 inches by inch, and 002 inches by inch each, laced inch apart, showing over them an increase of strength of nearly 50 cent. per In using this form of section, it makes no difference whether the load be placed wholly on the top of the vertical web, or on the lower flange; the result obtained in either case was the same.-Builder, 1845, vol. iii. p. 593. The results of some other experiments on this useful form of iren are given in the Engineer's Pocket-book for 1861, p. 205. The formula also applicable to the trough-shaped section, as N (fig. 613r.), as to the inverted Tee or shape M, taking the two vertical ribs to be equivalent to one rib of the same depth and double the thickness. The thickness of the horizontal and vertical parts of these girders should be equal, or nearly equal, so as to obtain as near an equality in cooling as possible. 1629a For K, Mixed beams; Flitch beams and double flitch beams :— These beams are composed of an iron plate (cast or wrought) placed betwen two pieces of fir timber, fig. 615s., or of a plate placed on each side of the solid timber beam, fig. 613t. These plates again may have a table or ! flange, as in the case of the single plate; or of a half flange, as in the case of the plates on each side of the beam. should be bolted, or otherwise secured together, to render them as homogeneous as possible. Hurst gives the formula All these Fig. 613r. ad C-6 I feet Fig. 6138. W tons is Fig. 613t. WI 8d = h; ;(Cb+30t) = W in cwts. Here t breadth of the one, or two, wrought iron flitches; b breadth of wood, both in inches; C, coefficient = 4 teak, 3 oak, 2:5 fir, and 20 elm. Fairbairn considers that "the addition of the timber on each side of the plate gives increased stiffness, and renders it less liable to warp under strain. It is called a sandwich beam." He states this beam "to be weak, comparing the results with those of the simple plate girder; and its elasticity, although considerable, is nevertheless so imperfect as to render it inadmissible for the support of great loads, whether proceeding from a dead weight, or one in motion over its surface. With riveted angles or flanges, the timbers on each side might have been useful in preventing lateral flexure, but they would not have contributed, in any great degree, to the vertical bearing powers of the beam." (Application, &c., p. 284-5.) Rolled flat irons can now be obtained about 13 to 14 inches deep, from inch to an inch in width, up to 30 feet in length, and for special cases somewhat longer. 1629. The method of trussing a beam is explained in CARPENTRY (par. 2021, et seq.). 1629w. The formulæ for finding the strength for examples IV. and V., fig. 675., are and 13+ Here length in feet; d depth in feet-both measured from the points of intersection of the stay, tension rod, and top beam; W load in tons uniformly distributed; h horizontal thrust on beam in tons; and s strain on inclined part of tension rod When the truss has more than one stay, h, l, and d will represent the same; and A tensile strain on the horizontal portion of the rod. The strain on the inclined tie rod will be F2+n22=s; n the number of times that the horizontal distance between the pier and the nearest stay is contained in 7. If any load be placed on the middle, the strain will be doubled. If any load be placed on each of the stays, then I will represent the distance of each loaded stay from the nearest pier; d depth as before; h horizontal thrust on the part next the pier; s tension on each of the inclined ties. Then h; and V+W=s. To resist the strains of the inclined tie rods with safety, allow an inch of Sectional area in the tie rod for every 5 tons of strain. The stay, being in compression, Should be calculated as a column capable of supporting the load if in the middle, or one half if distributed. The beam, though in compression, should be capable of supporting the in tons. WZ = 8. WI d = load between the stays, as a beam exposed to transverse strain, according to the rules before given. Tie rods, when exposed to great strains, are not generally of much value, because the iron strutches. 1629r. Mr. Cubitt experimented on an equal flanged cast iron girder, 27 feet long, 10 inches deep, and 4 inches broad across the flanges; the rods were 1 inch diameter. When the ends of the rods were placed above the beam, it was found to be weaker than having no rod at all. When the fastening was made at the upper end of the girder, and giving a distance to the rod of 64 inches below the girder instead of an inch, an increased stiffness was obtained of above a ton (Warr, p. 259). Some experiments are recorded in the Builder, 1857, on two beams of Dantzie timber, each 28 feet long, 14 inches with and without a tie rod. Barlow records an experiment (p. 158) on four beams, two being trussed similarly to the figure on plate xxxix. of Nicholson's Carpenters' New Guide. Mr. Cooper's experiments on trussed beams are given in the Builder, 1845, p. 612. For Trellis girders, another mode of trussing a beam, Fairbairn, p. 129, uses the same formula as for the plate girder F, but with the constant 60. For this, the student is referred to the Application &c. of Iron, enlarged edition, 1864. Other conditions than that of the Weight in the Middle. square, 1629y. To find the ultimate strength of a beam (section A or B), when a weight is placed somewhere between the middle and the end. RuleMultiply twice the length of the longer end, A, fig. 613u, by twice the length of the shorter end B, and divide the product by the whole length C, which will give the effective length to be used as the divisor for the calculation of strength under the conditions of the beam:-Thus say a beam is Fig. 613 2×10)x(2×5) =13-33 effective length; and (S=2548 cast iron)×2× 62 15 long Hurst puts it as, product of two lengths from each end. 13:33 =13,762-64 lbs. weight, 1629z. Barlow (p. 39-40) has stated a case where a beam has to support two equal weights between the points of support, FF', as at D and E, Example I. fig. 613, (ID+E, C,W then since IC=iC=I, and W=W', the general expression becomes ID+iE Ii × W=f. And if we suppose further ID=¡E, then it becomes simply ID.W=f 2 W W Fig. 613v. Now, if both weights act at the centre, it appears from the preceding investigation, that Ii. (2W) = } I¿.W = IC.W=f. Whence the strain in Ow the two cases will be to each other as ID to IC; and hence the following practical deduction:When a beam is loaded with a weight, and that weight is appended to an inflexible bar or bearing, as DE, in Ex. 2, the strain upon the beam will vary as the distance ID, or a the difference between the length of the beam and the length of the bearing; for th bearing DE being inflexible, the strains will be exerted in the points D and F, exactly i the same manner as if the bearing was removed, and half the weight hung on at each o these points. This remark may be worth the consideration of practical men in variou architectural constructions. He also puts the case of a beam, which, instead of being fixed a cach end, merely rests on two props, and extends beyond them on each side equal to half the distance, as Ex. 3: if the weights W W' were suspended from these latter points, eac equal to one-fourth of the weight W, then this would be double of that which would b necessary to produce the fracture in the common case; for, dividing the weight Wint four equal parts, we may conceive two of these parts employed in producing the strain fracture at E, and one of each of the other parts as acting in opposition to W and W and by these means tending to produce the fractures at F and F'. This is the cas which has been erroneously confounded with a former one (fixed at each end), but th distinction between them is sufficiently obvious; because here the tension of the fibres, the places where the strains are excited, are all equal; whereas in the former the midd one was double of each of the other two. 1630. Experiments are recorded in the Civil Engineer Journal, 1849, xi, page 44, o parallel bars of cast iron, 4 feet 8 inches long and 4 inches square, placed on tw |