Railway Engineering Test 1

Concept:

Actual cant or Equilibrium cant provided is given by 

\({e_{act}} = \frac{{G \times V_{avg}^2}}{{127 \times R\;}}\)

Where ,

G is gauge distance, 

V_avg or V_eq is average velocity or equilibrium velocity

Radius of the track

Calculation

Given,

R = 650 m, V_eq = 65 kmph

For Broad Gauge G = 1.676 m

\(e = \frac{{G{V^2}}}{{127R}} = \frac{{1.676 \times {{65}^2}}}{{127 \times 650}} = 8.58\;cm\)

∴ Equilibrium super elevation is 8.58 cm

Concept:

By Martins Formula

i) On a Transition Curves

a) For BG/MG track (≤ 100 km/hr)

\(\;{V_{max}} = 4.35 \times \sqrt {R - 67} \)

Where, V in kmph, R in m

b) For NG track

\({V_{max}} = 3.6 \times \sqrt {R - 60} \)

Where, V in kmph, R in m

ii) For High-Speed trains (> 100 kmph)

\(\;{V_{max}} = 4.58 \times \sqrt R \)

Where V in kmph, R in m

iii) On a Non Transitional curve

V_max = 80% of Speed of Transition Curve

Calculation:

Given,

V = 80 kmph

For Meter Gauge on A transitional curve 

\(\;{V_{max}} = 4.35 \times \sqrt {R - 67} \)

\(80 = 4.35 \times \sqrt {R - 67} \;\)

R = 405.22 m

Degree of curve is given by 

\(D = \frac{{1720}}{R} = \frac{{1720}}{{405.22}} = 4.24^\circ \)

∴D = 4.24° 

Concept

Grade compensation (GC) for BG = 0.04% per degree of curve

Calculation

Given,

Degree of curve D = 5° 

Ruling Gradient = 1 in 250 

So for 5° curve, compensation,

= 0.04 × 5 = 0.20%

∴ Ruling gradient \(= \frac{1}{{250}} \times 100 = 0.4{\rm{\% }}\)

∴ Allowable gradient \(= 0.4 - 0.2 = {\rm{\;}}\frac{{0.2}}{{100}} = \frac{1}{{500}}\)

∴ Allowable Gradient = 1/500

Concept:

Note: Length of transition curve is a factor of actual cant (e_act) or equilibrium cant (e_eq)is used 

As e_act is a factor of Equillibrium speed (V_avg) or Average speed (V_avg)

If V_max > 50 kmph

V_equ = 0.75 V­_max

Length of transition curve is maximum of the following:

1) L = 7.2 × e_a

2) L = 0.073 × D × V_max

3) L = 0.073 × e_a × V_max

Where, e_a is equilibrium cant and D is cant deficiency

Calculation:

Given,

Max permissible speed (V_max) = 100 kmph > 50 kmph

∴ V_eq = 0.75 × 100 = 75 kmph

Equilibrium cant \(\left( {{{\rm{e}}_{\rm{a}}}} \right) = \frac{{{\rm{G}} \times {\rm{V}}_{{\rm{eq}}}^2}}{{127{\rm{R}}}}\)

For BG track,

G = 1.676 m, and D = 75 cm, (speed ≤ 100 kmph)

\({{\rm{e}}_{\rm{a}}} = \frac{{1.676 \times {{75}^2}}}{{127 \times 570}} = 0.13{\rm{\;m}} = 13{\rm{\;cm}}\)

Length of transition curve is maximum of:

1) L = 7.2 × e_a = 7.2 × 13 = 93.6 m

2) L = 0.073 × D × V_max = 0.073× 7.5 × 100 = 54.45 m

3) L = 0.073 × e_a × V_max = 0.073 × 13 × 100 = 94.9 m

Concept

As different trains are moving at different speeds, we need to obtain an equilibrium speed

Given by weighted average 

\({V_{avg}} = \frac{{{n_1}{V_1} + {n_2}{V_2} + {n_3}{V_3} + {n_4}{V_4}}}{{{n_1} + {n_2} + {n_3} + {n_4}}}\)

Radius of the curve is given by

\(R = \frac{{1720}}{D}\)

Actual cant provided is given by 

\({e_{act}} = \frac{{G \times V_{avg}^2}}{{127 \times R\;}}\)

Grade compensation for Broad Gauge is 0.04 % per Degree

Calculations

Given,

Degree of curve (D) = 3° 

\(R = \frac{{1720}}{D}\)

= 1720/3 = 573.33 m

For Broad Gauge G = 1.676 m

Average speed or Equilibrium speed is 

\({V_{avg}} = \frac{{10 \times 50 + 10 \times 60 + 5 \times 40 + 4 \times 70}}{{10 + 10 + 5 + 4}}\)

= 54.48 kmph

Actual cant is given by 

\({e_{act}} = \frac{{G \times V_{avg}^2}}{{127 \times R\;}}\)

\({e_{act}} = \frac{{1.676 \times {{54.48}^2}}}{{127 \times 573.33\;}}\)

= 0.068 m

= 6.8 cm

ii) Grade compensation (G.C)

For broad gauge grade compensation = 0.04% per degree of curve

G.C = 0.04 × 3

= 0.12 %

= 0.12 × 10^-2

Concept:

For the Indian railways track

\({e_{th}} = \frac{{G \times V_{max}^2}}{{127\;R}}\)

Where,

e_th is the maximum allowable superelevation in (m) 

V_max is the maximum speed on the curve in kmph,

G is the gauge length of the track in m and

R is the radius of the curve in m where

D is the curvature of the curve in degree.

,\({\rm{R}} = \frac{{1720}}{{\rm{D}}}\), 

\({e_{th}} = {e_{act}} + CD\)

Where,

e_act is the actual or equilibrium superelevation 

CD is Cant deficiency

Calculation:

Given,

D = 3°, CD = 76 mm = 0.076 m

e_act = 10 cm = 0.1 m

As track is Broad Gauge G = 1.676 m

\({e_{th}} = {e_{act}} + CD\)

∴ e_th = 0.1 + 0.076 

= 0.176 m

Radius of Curvature  \({\rm{R}} = \frac{{1720}}{{\rm{D}}}\)

\(∴ {\rm{\;Radius\;of\;the\;curve}},{\rm{\;R\;}} = \frac{{1720}}{3}{\rm{m}} = 573.33{\rm{\;m\;}}\)

\({e_{th}} = \frac{{G \times V_{max}^2}}{{127\;R}}\)

\(0.176 = \frac{{1.676 \times V_{max}^2}}{{127 \times 573.33}}\)

V_max = 87.6 kmph

∴ The maximum speed of the train is approximately 87.6 kmph.

Important Point:

In the formula of maximum allowable super elevation \({e_{th}} = \frac{{G \times V_{max}^2}}{{127\;R}}\)  all the variables should be replaced in the particular unit mentioned above, otherwise the constant term ‘127’ is not valid.

Concept:

Shift of Transition curve is given by S = \(\frac{{L_s^2}}{{24R}}\)

The Railway Board has decided that on Indian Railways, the transition curve will normally be laid in the shape of a cubic parabola.

Equation of Cubic Parabola is given by y = \(\frac{{{x^3}}}{{6RL}}\)

Calculation:

Given,

L_s = 120 m, Shift (S) = 0.925 m

The radius of curve is

\(R = \frac{{{L^2}}}{{24 \times S}} = \frac{{{{120}^2}}}{{24 \times 0.925}}\)

= 648.64 m

Offset = \(\frac{{{x^3}}}{{6RL}}\)

At x = 40 m, offset is 

\(y = \frac{{{{\left( {40} \right)}^3}}}{{6 \times 684.64 \times 120}}\)

= 0.137 m = 13.7 cm