Raising outer edge for superelevation

Superelevation

$V=80$ km/h, $R=200$ m, mixed traffic.

• IRC (75% of speed, no friction): $e=\dfrac{V^2}{225R}=\dfrac{80^2}{225(200)}=0.142$, which exceeds the maximum 0.07 (IRC/NRS 2070 for plain/rolling terrain). Provide e = 0.07.
• Friction needed: $f=\dfrac{V^2}{127R}-e=0.252-0.07=0.182$, which is more than 0.15. A speed of 80 km/h is therefore not fully safe on this curve. The safe speed with $e=0.07$ is $\sqrt{127(200)(0.22)}=74.8$ km/h, so speed should be restricted to about 75 km/h (or $R\ge229$ m used).

The raising of the edge is worked out for $e=0.07$.

Width to be raised

Two lanes, $W=7.0$ m, extra widening ($l=6.1$ m): $W_m=\dfrac{2(6.1)^2}{2(200)}=0.186$ m, $W_{ps}=\dfrac{80}{9.5\sqrt{200}}=0.595$ m, $W_e\approx0.8$ m. Total width $B=W+W_e=7.8$ m.

(i) Rotation about the centre line

The outer edge rises and the inner edge falls by the same amount relative to the centre line:

$$h_{outer}=\frac{eB}{2}=\frac{0.07(7.8)}{2}=0.273\ \text{m}$$

The inner edge is lowered by 0.273 m below the centre line.

(ii) Rotation about the inner edge

The inner edge stays at its level and the outer edge is raised above it by the full amount:

$$h_{outer}=eB=0.07(7.8)=0.546\ \text{m}$$

The centre line is raised by 0.273 m.

(The camber is neglected here. In practice the outer half is first rotated flat, which adds the camber drop, and then rotated to the full superelevation.)

Answer: outer edge raised by about 0.27 m (centre-line rotation) or 0.55 m (inner-edge rotation) relative to the reference line.