Superelevation design for 150 m radius

Given

$V=60$ km/h, $R=150$ m, $e_{max}=0.07$, $f_{max}=0.15$ (IRC 73 / NRS 2070 approach).

(i) Equilibrium superelevation

Equilibrium superelevation is the value for which the centrifugal force is balanced entirely by the superelevation (no side friction, $f=0$):

$$e=\frac{V^2}{127R}=\frac{60^2}{127(150)}=0.189\ (=18.9\%)$$

(ii) Design superelevation (IRC method)

1. Superelevation for 75% of design speed, no friction: $e=\dfrac{V^2}{225R}=\dfrac{3600}{225(150)}=0.107$.
2. This exceeds $e_{max}=0.07$, so provide $e=e_{max}=\mathbf{0.07}$ (7%).
3. Check friction needed at full design speed:

$$f=\frac{V^2}{127R}-e=0.189-0.07=0.119$$

$f=0.119<0.15$, so the design is safe.

(iii) Maximum allowable speed if $e$ is not to exceed 0.07

With $e=0.07$ and $f=0.15$:

$$V=\sqrt{127R(e+f)}=\sqrt{127(150)(0.22)}=64.7\ \text{km/h}$$

Allowable speed is about 65 km/h.

(iv) Ruling radius for 60 km/h

$$R=\frac{V^2}{127(e+f)}=\frac{60^2}{127(0.07+0.15)}=128.8\ \text{m}$$

Answer: (i) 0.189; (ii) 0.07 (friction required 0.119); (iii) 64.7 km/h; (iv) about 129 m.