. Stresses in railway structures on curves . ble 4, we find that not only the maximumbending moments in both girders are made equal, but the maximum shears are also nearly equal, which fact is very desirable so far asthe girders only are concerned. But in securing such an arrangementthe spacing between the center lines of girders must be at feet instead of feet which is the required minimum spac-ing. This change of spacing of girders would no doubt increase thefloor beams and laterals in sections and lengths, the cost of whichmight offset the saving derived from having the two
. Stresses in railway structures on curves . ble 4, we find that not only the maximumbending moments in both girders are made equal, but the maximum shears are also nearly equal, which fact is very desirable so far asthe girders only are concerned. But in securing such an arrangementthe spacing between the center lines of girders must be at feet instead of feet which is the required minimum spac-ing. This change of spacing of girders would no doubt increase thefloor beams and laterals in sections and lengths, the cost of whichmight offset the saving derived from having the two girders beingmade of the same section. 67 CHAPTER BRIDGESArt. 12. Clearance. The formulae for determining the required clear width forthrough bridges have been found in Art. 9 of the last chapter; thatis, X = Z + Ma - MD - h| (I) and Y = | + - Mc + H| . (J) in which X and Y are the clear distances from the center line oftrack at the middle of the span to the outer and inner trusses re-spectively. The total clear width is X + Fig. 24. 68 For the present chapter, let us assume a through truss bridge of 200-foot span on a 5° curve. Also let the panel length be 25 feet, the height of truss be 34 feet, and the height of the center of gravity of trains be feet above the top laterals of the string ers and 10 feet above the bottom laterals of the trusses. Fig. 24 shows the type of truss and the general dimensions. The other data are taken from Table A of Art. 5 as follows: Radius of curve, r = ft. Centrifugal force coefficient, q = .1315 Super-elevation of outer rail, s = .6496 ft. Distance between center line of track and curve of center of gravity of train, c = .6575 ft. The middle ordinates for chords A, B, G are found from the approximate formula M = thus For A = 80 Ma - .6979 ft. B = 60 Mb = .3926 ft. C 5 175* Mc = ft. Further, we have w = 14, h = 4, and H = 14. Hence from Equation (I) we find X = ~ * .6979 - .3926 - 4 x c - 7 + .6979 - .392
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