. Stresses in railway structures on curves . ges of the girders is945,000/ - 49,000 lbs., being in tension for the outer and incompression for the inner girder. 64 (2) Maximum Shears. The influence ordinates for shears are computed as follows 56 ?anel point B 2 = ■£ e = 6x1 a. 25 5 6x19 . 25 - 2 2 3x19. 25 2 ( - 2 x .9513+ 2 x .8613 Panel point C Z = | e - Outer Girder: Ord. c r7_^ -—( - 2 x .5980 + 2 x ) = .7133 Inner Girder: Ord. = g & 19 2&( + 2 x .5980) = .7081 Panel point D 2 = \ e = tL_ 2 Outer Girder: Ord. ^j^ 19 •25 - 2 x .4802 + 2
. Stresses in railway structures on curves . ges of the girders is945,000/ - 49,000 lbs., being in tension for the outer and incompression for the inner girder. 64 (2) Maximum Shears. The influence ordinates for shears are computed as follows 56 ?anel point B 2 = ■£ e = 6x1 a. 25 5 6x19 . 25 - 2 2 3x19. 25 2 ( - 2 x .9513+ 2 x .8613 Panel point C Z = | e - Outer Girder: Ord. c r7_^ -—( - 2 x .5980 + 2 x ) = .7133 Inner Girder: Ord. = g & 19 2&( + 2 x .5980) = .7081 Panel point D 2 = \ e = tL_ 2 Outer Girder: Ord. ^j^ 19 •25 - 2 x .4802 + 2 x ) - .5413 Inner Girder: Ord. = gxi^ 25(19-25 + 2 x -4802) r .5249Panel point E Z 3 e = -5980 Outer Girder: Ord. = —-±—-( - 2 x .5980 + 2 x ) = .3567 Inner Girder: Ord. - —J__( + 2 x .5960) - .3540 Panel point F Z - - e = . o Outer Girder: Ord. = -■ ■ ( - 2 x .9513 +2 x ) s .1723 Inner Girder: Ord. = ^ ^-( + 2 x .9513) = .1931. 65 A 5 C D £ f 5. Fig. 23. The influence lines are shown in Jig. 23, (a) heing forgirders on straight track, and (o) and (c) being for the outer andinner girders on curved track. The area of influence lines for the outer girder is15 (.8613 + .7133 + .5413 + .3567 + .1723): 15 x = that for the inner girder is A2 = 15 (.9157 + .7081 + .5249 + .3540 + .1931)= 15 x = 66 Using an equivalent uniform load of 3490 lbs. per foot ofgirder, we have the maximum shear in the outer girder Y1 - x 3490 = 138,600 the maximum shear in the inner girder isV£ = x 3490 - 141,200 lbs. The shears and moments found above are now tabulated in Table 4. TABLE 4. Shears, Moments and direct Stressesin a Through Plate Girder Bridge of on a 6°Curve. Shearsinlbs. Momentsin ft. lbs. Direct Stressesin Bottom Flangein lbs. Bridge on Tangent Each Girder 131,0U0 3,337,000 0 Bridge on Curve Outer Girder 138,600 3,557,000 + 49,000 Inne
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