Question 2 of 7: Horizontal Curve Design and Superelevation Development

Question 2: Horizontal Curve Design and Superelevation Development (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. A circular curve of acceptable radius and superelevation for 80 km/h, and the four stations that define the superelevation development when the pavement is rotated about its centreline with no spiral.

Approach. Fix the smallest radius permitted for 80 km/h at e_max = 0.04 from the supplied table, adopt the next tabulated radius as the design radius and read its superelevation and spiral parameter; convert the spiral parameter into the superelevation runoff length, derive the tangent runout from the ratio of crown to full superelevation, then chain forward and backward from the PI, placing 0.70 of the runoff on the tangent as TAC practice requires when no spiral is used.

1. Part — minimum radius at 80 km/h. The point-mass relation gives the smallest radius that can be driven at the design speed with the maximum superelevation and the design side-friction factor f = 0.14: $$R_{\min} = \frac{V^2}{127(e_{\max}+f)} = \frac{80^2}{127(0.04+0.14)} = \frac{6400}{22.86} = 280.0\ \text{m}$$ The supplied Table B.3.1.4a confirms this exactly: the 80 km/h column terminates at min R = 280.
2. Select the design radius and read its superelevation. Designing at the absolute minimum leaves no tolerance, so the next tabulated radius is adopted. Entering the table at R = 300 m in the 80 km/h column returns a superelevation of e = 0.040 m/m and a minimum spiral parameter A = 135 m for a two-lane pavement: $$\boxed{R = 300\ \text{m}, \quad e = 0.040\ \text{m/m}, \quad A = 135\ \text{m}}$$
3. Tangent length and arc length of the circular curve. For a simple circular curve of deflection Δ, $$T = R\tan\frac{\Delta}{2} = 300\tan 12.5^\circ = 300(0.221695) = 66.508\ \text{m}$$ $$L_c = \frac{\pi R \Delta}{180} = \frac{\pi (300)(25)}{180} = 130.900\ \text{m}$$
4. Chainage of the PI and of the tangent points. With 1000 m stations, station 30 + 000.000 is chainage 30 000.000 m. Subtracting the tangent length places the PC, and adding the arc places the PT: $$\begin{aligned} \text{Ch}_{PC} &= 30\,000.000 - 66.508 = 29\,933.492\ \text{m} \\ \text{Ch}_{PT} &= 29\,933.492 + 130.900 = 30\,064.391\ \text{m} \end{aligned}$$
5. Superelevation runoff length from the spiral parameter. The note beneath the table defines the spiral length as $L = A^2/R$. Although this curve carries no spiral, that length remains the distance over which the outer edge is rotated from a level cross-section to full superelevation: $$L_r = \frac{A^2}{R} = \frac{135^2}{300} = \frac{18\,225}{300} = 60.75\ \text{m}$$ The implied relative gradient of the outer edge against the centreline is $0.150/60.75 = 0.00247$, i.e. 1 in 405 — flatter than the 1 in 200 maximum for an 80 km/h two-lane road, so the runoff is comfortably long enough.
6. Tangent runout length. The runout removes the adverse crown at the same edge-rotation rate used in the runoff, so it is proportional to the crown slope: $$L_t = L_r\,\frac{e_{NC}}{e_d} = 60.75\left(\frac{0.020}{0.040}\right) = 30.375\ \text{m}$$
7. Distribute the runoff about the PC. With rotation about the centreline and no spiral, TAC places 0.70 of the runoff on the tangent and 0.30 within the circular curve, so that the pavement reaches roughly 0.7e at the PC: $$\begin{aligned} 0.70 L_r &= 0.70(60.75) = 42.525\ \text{m} \\ 0.30 L_r &= 18.225\ \text{m} \end{aligned}$$
8. Station (b) — junction of tangent to transition length. This is the point at which the outer half of the pavement has just been brought level, the runout ends and the runoff begins. It lies 0.70L_r ahead of the PC: $$\text{Ch} = 29\,933.492 - 42.525 = 29\,890.967\ \text{m} \Rightarrow \boxed{29+890.967}$$
9. Station (a) — beginning of the tangent runout. One further runout length back along the tangent, where the pavement still carries its full normal crown: $$\text{Ch} = 29\,890.967 - 30.375 = 29\,860.592\ \text{m} \Rightarrow \boxed{29+860.592}$$
10. Stations (c) and (d) — the circular curve. These are the PC and PT computed in step 4: $$\boxed{\text{PC} = 29+933.492, \quad \text{PT} = 30+064.391}$$ Full superelevation is reached 18.225 m beyond the PC, at chainage 29 951.717 m, and is held until 18.225 m before the PT, at 30 046.166 m, after which the sequence reverses symmetrically.
11. Edge elevations for the diagram. Rotating about the centreline, the outside edge moves from $-0.020(3.75) = -0.075$ m at normal crown, through 0 at the end of the runout, to $+0.040(3.75) = +0.150$ m at full superelevation, a total rise of 0.225 m; the inside edge falls from −0.075 m to −0.150 m over the runoff only. Plotted at a horizontal scale of 1:500 and a vertical scale of 1:5 the diagram is exaggerated 100 times vertically, which is what makes the small edge movements legible.