Question

A BG main line curve of 2.76° and a 1:12 turnout takes out in contrary flexure for a branch line which has a design speed of 30 kmph. The permissible cant on the main line is considering inner to inner distance of the rail as gauge.

• A. 56.8 mm
• B. 65.2 mm
• C. 61.2 mm
• D. 88.8 mm

Answer 1

The correct option is C 61.2 mm

Degree of turnout = $ta{n}^1\left(1/12\right)$ = 4.76°

Degress of main line = 2.76°

Degree of branch line = 4.76 - 2.76 = 2°

Radius of branch line = 1720/D = 1720/2 = 860m

Cant required for branch line = $\frac{G{V}^2}{127R}=\frac{1676\times {30}^2}{127\times 860}=13.8mm$
Maximum permissible cant deficiency = 75 mm for BG track
Since negative cant is provided on branch track,
Cant for main track = 75 - 13.8 = 61.2 mm.