. Stresses in railway structures on curves . ausethe train moving on a tangent instead of moving on the curved track,and hence there must be a centripetal force tc keep that train movingon its path. If the two rails of the trac.*. were laid on the samelevel transversely, this centripetal force could only be furnishedby the pressure of the wheel-flanges against the outer rail. Thiscondition is very objectionable and undesirable. According to thepresent practice, the cuter rail of the track is so elevated abovethe inner that the reaction of the rails against the wheels of thetrain shall contain
. Stresses in railway structures on curves . ausethe train moving on a tangent instead of moving on the curved track,and hence there must be a centripetal force tc keep that train movingon its path. If the two rails of the trac.*. were laid on the samelevel transversely, this centripetal force could only be furnishedby the pressure of the wheel-flanges against the outer rail. Thiscondition is very objectionable and undesirable. According to thepresent practice, the cuter rail of the track is so elevated abovethe inner that the reaction of the rails against the wheels of thetrain shall contain a horizontal component equal to the requiredcentripetal force. In Fig. 2, let 0 be the center of gravity of the 0 A represents the centrifugal force F, and 0 G, the weight of 5 the train; and 0 B, resultant of 0 A and 0 C, is perpendicular tothe plane of the tops of rails and passes through the center line ofthe track; then B 0 will be the reaction of the rails against thewheels of the train, and B C, the required centripetal force. Fig. 2. ( - 0 A = F). P M is the distance between the centers of rails andLI N is the required super-elevation of the outer similar triangles 0 B C and P M H, 0 C : B C - P XT : M N • » . - b C. P U * * - 0~G Now, P U is the horizontal distance between the centers ofthe rails and is slightly less than P M. Hence approximately we maywrite Jf If ; B 0 • PM- 0 0 6 Let 8 = M K = super-elevation of outer rail. And t - ? M * distance between rail centers. v2 Since BC = OA = F = -— W g r 0 c = w . s = t -—g r For standard gauge of 4- 8t> and a 100-lb. rail, thedistance between rail centers is t = 4-8| 4 2|= 4 .229 = feet. For convenient use in the following analysis, t will be taken as 2 feet. The values of s = t^— for curve of different degreesare shown in the table in Art. 5. Art. 4. Center of Gravity of Train. The vertical line through the center of gravity of a trainmoving on a tangent always intersects
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