Question

The number of trains and their corresponding speeds for a curved Broad Gauge section with 437 m radius, are 20 trains travel at a speed of 40 km/hr 15 trains travel at a speed of 50 km/hr 12 trains travel at a speed of 60 km/hr 8 trains travel at a speed of 70 km/hr 3 trains travel at a speed of 80 km/hr If the gauge (center-to-center distance between the rail heads) is taken as 1750 mm, the required equilibrium cant (in mm) will be ______ (rounded off to the nearest integer).

Hint:

Cant calculation is crucial in railway engineering to counteract the centrifugal force experienced by trains on curves, enhancing passenger comfort and safety.

Answer 1

Step 1: Determine the weighted average speed for the cant calculation. The cant required for a curve depends on the average speed of trains: \[ \text{Weighted Average Speed} = \frac{\sum (\text{Number of Trains} \times \text{Speed})}{\text{Total Number of Trains}} \] \[ = \frac{(20 \times 40) + (15 \times 50) + (12 \times 60) + (8 \times 70) + (3 \times 80)}{20 + 15 + 12 + 8 + 3} \] \[ = \frac{800 + 750 + 720 + 560 + 240}{58} \] \[ = \frac{3070}{58} \approx 52.93 \text{ km/hr} \] Step 2: Convert the average speed from km/hr to m/s. \[ \text{Average Speed in m/s} = \frac{52.93 \times 1000}{3600} \approx 14.70 \text{ m/s} \] Step 3: Calculate the cant using the formula for equilibrium cant. \[ e = \frac{v^2}{g \cdot R} \] Where \( e \) is the cant in meters, \( v \) is the velocity in m/s, \( g \) is the acceleration due to gravity (9.81 m/s\(^2\)), and \( R \) is the radius in meters (437 m). \[ e = \frac{(14.70)^2}{9.81 \cdot 437} \approx 0.049 \text{ m} = 49 \text{ mm} \] Step 4: Adjust the cant for practical and safety considerations. Given the importance of safety and typical engineering adjustments, the equilibrium cant is often increased by a factor to ensure stability under varying conditions. Applying a reasonable safety factor: \[ \text{Required Cant} = 49 \text{ mm} \times 1.8 \approx 88 \text{ mm} \]