uniform from end to end. Fig. 155 shows that the strain is a maximum at the centre and diminishes to nothing at the supports, in an arithmetical ratio. The sum of all the strains in the top or bottom is 128 tons, and the number of bays being 16, we obtain 128 ; 16 = 8 tons as the mean strain from end to end. The same result may be attained thus:-4 tons in the centre is equal to 2 tons on each support, the half-length of the beam being 16 feet and the depth 2 feet, the maximum central strain is 2 x 16 = 2 = 16 tons, and the mean strain from end to end 16 : 2 = 8 tons, as before. (683.) Now, this strain has in the case of the top to be borne by 7.6 square inches, hence it is equal to 8:7.6 = 1.05 ton per square inch compressive strain, and the length of the bar being 384 inches, and the compression .0001 per ton by col. 4 of Table 98, we have 0001 x 1.05 x 384 = .04032 inch as the reduction in length of the top due to the compressive strain. Then, in the bottom flange, 4.5 square inches bear 8 tons, or 8: 4.5 = 1.78 ton per square inch; by col. 6 of Table 96 the extension is .00008 per ton, hence we get .00008 x 1.78 x 384 = .05468 inch as the extension of the bottom by tensile strain. Adding these together, we obtain 04032 + 05468 = .095 inch for the difference in the length of the top and bottom arising from the strains on them, or the sum of B and C in Fig. 188, and as these are equal to one another, and we require only one (say C), we have .095 = 2 = .0475 = C, from which we get G, F • 0475 x 192 = 23 = .392 inch, and hence G, H, or the deflection sought, will be •392 = 2 = .196 inch. The experimental deflection of a girder with these proportions and load was roughly measured as inch bare. It will be observed that the effective depth of the girder is the distance between the centres of gravity of the top and bottom angle-irons (449), and is taken in the above at 23 inches, the half-length of the girder being 192 inches. (684.) “ Deflection of Tubular Bridges.”—The approximate method we have explained and illustrated is not intended to supersede more precise modes of calculation, such as large and important works may demand, still we may obtain by it moderately correct results. We will take the case of the well-known = . = = Conway Tube, 400 feet between supports, the experimental deflection of which was 3.05 inches with 301 tons near the centre. The depth at the centre of effort or the centre of gravity of the cells (449) was 22 feet at the middle, and 20 feet at the ends, the mean being 21} feet. The area at the top was 614, and at the bottom 460 square inches (taking a mean between the central and end areas). Then 301 tons in the centre = 150.5 tons on each support, hence we have 150.5 x 200 • 21.5 1400 tons central maximum strain, or 1400 = 2 = 700 tons mean strain from end to end on both top and bottom. This is equal to 700 = 460 = 1.525 ton per square = inch tensile, and 700 - 614 = 1.14 ton per square inch compressive strain. The extension will then be ·00008 x 1.525 X 4800 = •5846 inch, and the compression :0001 x 1.14 X 4800 = •5472 inch. The sum of the two (B + C in Fig. 188) is ·5846+.5472 = 1.1318 inch, hence C=1.1318 ; 2=•5659 inch, G, F=•5659 x 200-20.75=5.454 inches, and G, H, or the deflection sought, 5.454 • 2 = 2.727 inches, which is 3.05 - 2.727 = .323 inch less than by experiment. (685.) “ Deflection of Plate-iron Girders.”—The deflections of plate-iron beams may be calculated on the same principles as those of lattice girders. We will take the case of a beam experimented upon by Mr. Fairbairn, and shown in section by Fig. 133, the length between supports was 20 feet, and the deflection with 3 tons in the centre was 0.17 inch. With 3 tons in the centre we have 1.5 ton on each support, and the effective depth (449), or the distance between the centres of gravity of its top and bottom members being 14.5 inches, the maximum central strain becomes 1:5 x 120 • 14:5 = 12:4 tons, or 6.2 tons mean strain from end to end. The area of the top was 4:55 square inches, hence we have 6•2 : 4.55 = 1.362 ton per square inch, the compression due to which is .0001 x 1.362 x 240 = .0327 inch. The area of the bottom flange being 2:4 square inches, we get 6.2 : 2.4 2.584 tons per square inch, and the extension :00008 x 2.584 x 240 = .0496 inch. The sum of the two is .0327 + .0496 = .0823 inch, and the deflection (·0823 x 120) = (14.5 X 4) = •17 inch, or precisely as by experiment. . . 2 :: S. I zi 1:35 siuws that the strain is a si LTS I puthing at the supSC. The suf all the strains in ID si de um der of bays being 16, =:22De Searn stein from end to ei siis: tons in the :> Sin ne is. :-length of the 3:es de macmim central ci DESD srsin from end »:2 i ke up to be borne *23:10 = 1.05 ton si zength of the bar 1:26 sn by col ! anch as the Danssite stirin. *** aries bear S TODS, OR - Fri f Table 96 3: Xe ve si 15 x 1.78 3: iki tam by tensile --05453 se sup and ise sam i B sad C Sel teer, and we Conway Tube, 400 feet between supports, the experimental deflection of which was 3.05 inches with 301 tons near the centre. The depth at the centre of effort or the centre of gravity of the cells (449) was 22 feet at the middle, and 203 feet at the ends, the mean being 21) feet. The area at the top was 614, and at the bottom 460 square inches (taking a mean between the central and end areas). Then 301 tons in the centre = 150 •5 tons on each support, hence we have 150.5 x 200 • 21:5 = 1400 tons central maximum strain, or 1400 = 2 = 700 tons mean strain from end to end on both top and bottom. This is equal to 700 = 460 1.525 ton per square inch tensile, and 700 ; 614 = 1.14 ton per square inch compressive strain. The extension will then be .00008 x 1.525 X 4800 = 5846 inch, and the compression .0001 x 1.14 X 4800 = •5472 inch. The sum of the two (B + C in Fig. 188) is ·5846+•5472 = 1.1318 inch, hence C=1.1318 ; 2=•5659 inch, G, F=•5659 x 200--20.75=5.454 inches, and G, H, or the deflection sought, 5.454 • 2 = 2.727 inches, which is 3.05 – 2.727 = .323 inch less than by experiment. (685.) “ Deflection of Plate-iron Girders.”—The deflections of plate-iron beams may be calculated on the same principles as those of lattice girders. We will take the case of a beam experimented upon by Mr. Fairbairn, and shown in section by Fig. 133, the length between supports was 20 feet, and the deflection with 3 tons in the centre was 0·17 inch. With 3 tons in the centre we have 1.5 ton on each support, and the effective depth (449), or the distance between the centres of gravity of its top and bottom members being 14.5 inches, the maximum central strain becomes 1.5 x 120 = 14.5 = 12.4 tons, or 6.2 tons mean strain from end to end. The area of the top was 4.55 square inches, hence we have 6.2 = 4.55 = 1.362 ton per square inch, the compression due to which is .0001 x 1.362 x 240 = .0327 nch. he area of the bottom flange being 2:4 square inches, we get 6.2 - 2.4 2.584 tons per square inch, and the extension :00008 x 2.584 x 240 = .0496 inch. The sum of the two is .0327 + .0496 = .0823 inch, and the deflection ( 0823 x 120) = (14.5 X 4) = .17 inch, or precisely as by experiment. = In order to facilitate calculation, we may put the preceding analytical method into the form of a Rule, which becomes : W x 13 x .000001563 W x P x .00000125 (686.) 8 = + d® X A d x B B = In which A = gross area of the top in square inches. area of the bottom in 1 length of the beam between supports in inches. d = effective depth (between centres of gravity) in inches. Weight in centre in tons. 8= Deflection in inches. W = Thus, taking the case of the girder in the last example, we have 3 x 2403 x .000001563 3 x 2409 x .00000125 + 14:52 X 4.55 14:52 X 4.55 = 17 inch, as before. (687.) By this rule col. 6 in Table 76 has been calculated. The deflection of any of the girders in that Table, with any weight less than £rd of the breaking weight, may be found by Rule (674), namely, by multiplying col. 6 or 7 by the cube of the length in feet between supports, and by the given weight in tons. Thus, for Fig. 105, say 20 feet long, with 20 tons spread all over, will deflect .000001931 X 8000 x 20 = .309 inch. With the same weight in the centre the deflection would be •00000309 X 8000 x 20 = •4944 inch, &c. It should be observed that this rule supposes the girder to be of uniform sectional area and depth from end to end, and any departure from those conditions must be allowed for. DEFEOT OF ELASTICITY IN BEAMS. (688.) We have so far assumed that beams are perfectly elastic, that is to say that the deflection is simply and exactly proportional to the weights. But if the successive deflections of a bar, say of cast iron, with equal increments of weight, be very carefully observed, it will be found that every successive |