Question

Example 4 A BG branch line track takes off as a contrary flexure through 1° turnout from a main line track of a 3º curvature. Due to the turnout, the maximum permissible speed on the branch line is 30 km/h. calculate the negative superelevation to be provided on the branch line track and the maximum permissible speed on the main line track (when it takes off from a straight track).

Answer 1

Sol Sol Given that the branch lino for missible speed on (VBT) permissible 30km/hour For Branch track : OBE ( given) Cant for maximum permissible Speed GO2 e = 127 R for @= Cant maximum permissible spent (in mm) Maximum permissible speed G= 1676 (for BG V = R= Radius Track) R = 1750 0 R = 1750 17 So meter - 2 1676 x 30 50 e- = 6.8 mm 127 X1750 .8 mm le-66 To Round off to nearest (5mm) = lomm es lomm

The Value of for nog aliva Super Clevation branch line track's X = e-Cd Cd Jx Cant daffienciy For Broad Gauge (8.G) | Cd = 75 mm negative cont / Supercloration .65mm Ya 10-75 px=-65 man Then the Super elevation of the main track line is 65 mm, It is same Super cloration of the branch line track but as the XBT XMT So XBT xmT - (-65) xmt X MT +65mm XMT negative cant for main trock.

for Main track: Cact IMT + Cd 65 + 75 I Cact 140mm MT RMT 583.33 meter 3° C given) 1750 3 The maximum possible speed on main line track, the G22 lemt). act. 127R 1676x22 140 = 127 x 583.33 162 140 x127 X 583.33 16.76 6188.309 و after Solving we get V - 78.66Stemin

for check The maximum permissible Speed Calculated as va 0.27 Jica + Cij XR where Ca= negative cant VE Cd = Cant deficiency 0.27 (65+75)¥ 583:3 77.158 kmln. Ans lo V = There fort the maximum permissible speed on the main line track is 77.16 km/h