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Transition curve 220 m, pavement 7 m

Short answer

Length of transition curve is 60 m, shift is 0.68 m. Superelevation limited to 0.07 governs.

Calculate the length of transition curve and shift, if the design speed is 60 kmph. The radius of circular curve is 220 m. An allowable rate of change of centrifugal acceleration is 60 cm/sec³. Allowable rate of change of superelevation is 1 in 120. The width of the pavement is 7.00 m. Assume rotation of Pavement about the inner edge.

Superelevation

e=V2225R=602225(220)=0.0727e=\dfrac{V^2}{225R}=\dfrac{60^2}{225(220)}=0.0727, which exceeds the maximum 0.07 (IRC/NRS), so e=0.07e=0.07. Friction needed: V2127R−e=0.059<0.15\dfrac{V^2}{127R}-e=0.059<0.15, so it is safe.

Length of transition curve

  1. Rate of change of centrifugal acceleration (C=0.6C=0.6 m/s³, V=60/3.6=16.67V=60/3.6=16.67 m/s):
Ls=V3CR=16.6730.6(220)=35.07 mL_s=\frac{V^3}{CR}=\frac{16.67^3}{0.6(220)}=35.07\ \text{m}
  1. Rate of introduction of superelevation (1 in 120), pavement rotated about the inner edge, so the outer edge must be raised by eW=0.07×7.0=0.49eW=0.07\times7.0=0.49 m relative to the inner edge:
Ls=N e W=120(0.07)(7.0)=58.8 mL_s=N\,e\,W=120(0.07)(7.0)=58.8\ \text{m}

The greater value governs: adopt L_s = 60 m.

Shift

s=Ls224R=60224(220)=0.682 ms=\frac{L_s^2}{24R}=\frac{60^2}{24(220)}=0.682\ \text{m}

Answer: length of transition curve = 60 m; shift = 0.68 m. (IRC empirical value 2.7V2/R=44.22.7V^2/R=44.2 m is smaller, so it does not govern.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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