Transition curve and edge elevations, hilly terrain

Given: $R = 325$ m, $V = 60$ km/h, two-lane road in hilly terrain with heavy rainfall, rate of introduction of superelevation 1 in 60, centre-line levels 312.23 m (straight) and 312.9 m (circular curve), parabolic camber 2.5%.

Assumptions: carriageway $W = 7.0$ m; extra widening is neglected for edge levels; the pavement is rotated about the centre line.

Step 1: Superelevation

$$e = \frac{V^2}{225R} = \frac{60^2}{225\times325} = 0.0492 \approx 5\%$$

Check: $f = \dfrac{60^2}{127\times325} - 0.05 = 0.037 < 0.15$, OK. ($e_{max}=10\%$ in hills, and not more than 7% in snowy areas.)

Step 2: Length of transition curve ($v = 16.67$ m/s)

• Centrifugal acceleration: $C = \dfrac{80}{75+60} = 0.593$ m/s³, $L_s = \dfrac{v^3}{CR} = \dfrac{16.67^3}{0.593\times325} = 24.0$ m
• Superelevation rate 1 in 60: the outer edge rises from $-nW/2 = -0.0875$ m (below the crown) to $+eW/2 = +0.175$ m, a rise of 0.2625 m: $L_s = 60\times0.2625 = 15.75$ m
• IRC empirical (mountainous/steep terrain): $L_s = \dfrac{V^2}{R} = \dfrac{60^2}{325} = 11.1$ m (the formula $2.7V^2/R$ applies only to plain and rolling terrain)

Adopt the largest, rounded up: $L_s = 25$ m.

Step 3: Outer edge elevations

Straight section (centre line 312.23 m): with a parabolic camber of 2.5% on a 7.0 m carriageway, the drop from crown to edge $= n\,W/2 = 0.025\times3.5 = 0.0875$ m.

$$RL_{outer} = 312.23 - 0.0875 = \mathbf{312.14\ m}$$

Circular curve section (centre line 312.9 m): full superelevation of 5%, with the pavement rotated about the centre line, so the outer edge is raised $eW/2 = 0.05\times3.5 = 0.175$ m.

$$RL_{outer} = 312.90 + 0.175 = \mathbf{313.08\ m}$$

(The inner edge is at $312.90 - 0.175 = 312.73$ m.)

Answer: $e = 5\%$, $L_s = 25$ m; outer edge levels: 312.14 m on the straight and 313.08 m on the circular curve.