16-Civ-B7 Transportation Planning and Engineering · December 2015

Question 2 of 8: Superelevation Design from Coordinate Geometry

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 98-Civ-B7 Highway Engineering, National Examinations, December 2015. Three hours, open book, any non-communicating calculator. Eight questions of equal value (20 marks each); five solutions constitute a complete paper and only the first five in the answer book are marked. Note 1 invites the candidate to state any assumption made about an ambiguous input, and Note 2 permits any datum that is required but not given to be assumed. All eight questions are solved here, because the set is a study resource rather than a timed attempt.

Reference texts. Garber & Hoel, Traffic and Highway Engineering, 5th ed. (geometric design, sight distance, pavement design); Transportation Association of Canada, Geometric Design Guide for Canadian Roads (superelevation and spiral tables — the paper's Table 2.1.2.5 is TAC page 2.1.2.12); AASHTO, A Policy on Geometric Design of Highways and Streets (Green Book) for runoff distribution and relative-gradient limits; AASHTO, Guide for Design of Pavement Structures (1993) for the flexible pavement equation and layer/drainage coefficients; Asphalt Institute MS-2, Asphalt Mix Design Methods and Mamlouk & Zaniewski, Materials for Civil and Construction Engineers, for mixture volumetrics and binder grading.

Check: assumptions carried through this paper. Under the paper's own Note 2 the following values are assumed and stated where used: the AASHTO maximum relative gradient (0.50 % at 80 km/h) and the 70 % / 30 % split of superelevation runoff either side of the PC for two lanes rotated (Question 2); a truck factor of 0.52 for all trucks on a rural Interstate and a lane-distribution factor of 0.70 for three lanes in one direction (Question 6); and a downhill 2 % ramp grade in Question 4, since the freeway is elevated above the local street. Each is flagged again at the point of use with the sensitivity of the answer to it.

Question 2: Superelevation Design from Coordinate Geometry (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. A three-tangent alignment defined by four coordinate pairs, joined by two simple curves without spirals, on an urban undivided collector with a design speed of 80 km/h.

Alignment and cross-section data
Item	Value
BOP (N, E), station 0+000	765.978, 765.978
PI 1 (N, E), radius	1000.000, 1000.000; R₁ = 509.296 m
PI 2 (N, E), radius	1000.000, 1500.000; R₂ = 408.105 m
EOP (N, E)	1200.000, 1700.000
Design speed / class	80 km/h, urban collector undivided
Lanes, width	4 (2 each way), 3.5 m
Maximum superelevation	$e_{max} = 0.04$ m/m (TAC Table 2.1.2.5)
Normal cross slope	0.020 m/m; rotation about the centreline; no spiral

Find. For each curve, the stations of the normal-crown point (NC), the removal-of-adverse-crown point (RC), the PC, the beginning of full superelevation (FS) and the PT, together with the cross slope carried at each of those points.

Plan of the alignment. The two 45° deflections are of opposite sense, so the pavement rotates from a right-hand superelevation on Curve 1 to a left-hand superelevation on Curve 2.

Approach. Recover the tangent bearings and lengths from the coordinates, obtain the deflection angles and hence the curve geometry and stationing; read the design superelevation from the TAC table at each radius; compute the superelevation runoff and tangent runout lengths; and distribute the runoff about the PC in the conventional 70 % / 30 % proportion.

Tangent bearings and lengths from the coordinates. Differencing northings and eastings, $$\text{BOP} \to \text{PI}_1:\ \Delta N = \Delta E = 234.022\ \text{m} \ \Rightarrow\ \text{N }45^\circ\text{ E},\ \ell_1 = 330.957\ \text{m}$$ $$\text{PI}_1 \to \text{PI}_2:\ \Delta N = 0,\ \Delta E = 500.000 \ \Rightarrow\ \text{due east},\ \ell_2 = 500.000\ \text{m}$$ $$\text{PI}_2 \to \text{EOP}:\ \Delta N = \Delta E = 200.000 \ \Rightarrow\ \text{N }45^\circ\text{ E},\ \ell_3 = 282.843\ \text{m}$$ Both deflections are therefore $\Delta = 45^\circ$ — the first to the right, the second back to the left.
Curve geometry. With $T = R\tan(\Delta/2)$ and $L_c = R\,\Delta$ in radians, $$T_1 = 509.296\tan 22.5^\circ = 210.957\ \text{m},\qquad L_{c1} = 509.296\left(\tfrac{\pi}{4}\right) = 400.000\ \text{m}$$ $$T_2 = 408.105\tan 22.5^\circ = 169.043\ \text{m},\qquad L_{c2} = 408.105\left(\tfrac{\pi}{4}\right) = 320.525\ \text{m}$$ The radii have evidently been chosen so that the curve lengths and the tangent points fall on round stations.
Stationing along the final route. Chaining from the BOP and deducting each tangent distance as the route leaves the straight, $$\text{PC}_1 = 330.957 - 210.957 = \boxed{0+120.000}, \qquad \text{PT}_1 = 120.000 + 400.000 = \boxed{0+520.000}$$ $$\text{PC}_2 = 520.000 + (500.000 - 210.957) - 169.043 = \boxed{0+640.000}, \qquad \text{PT}_2 = \boxed{0+960.525}$$ and the alignment ends at station 1+074.325. A tangent of exactly 120.000 m separates PT₁ from PC₂; that length will govern whether the two superelevation transitions fit.
Design superelevation from TAC Table 2.1.2.5. Entering the 80 km/h, $e_{max} = 0.04$ column, the tabulated radii bracket both curves: $R = 600$ m gives $e = 0.029$ and $R = 500$ m gives $e = 0.033$; $R = 500$ m gives 0.033 and $R = 400$ m gives 0.037. Interpolating linearly on radius, $$e_1 = 0.033 - \tfrac{9.296}{100}(0.004) = 0.0326 \to \boxed{e_1 = 0.033}$$ $$e_2 = 0.037 - \tfrac{8.105}{100}(0.004) = 0.0367 \to \boxed{e_2 = 0.037}$$ Both interpolated values round to the tabulated entry at the next smaller radius, so the table can be read directly and the ambiguity disappears.
Superelevation runoff length. Rotation is about the centreline of an undivided four-lane pavement, so each half of the section rotates two lanes: $n_1 = 2$ and $w n_1 = 7.0$ m, with the AASHTO multilane adjustment $b_w = [1 + 0.5(n_1-1)]/n_1 = 0.75$. Taking the maximum relative gradient at 80 km/h as $\Delta_{max} = 0.50\ \%$, $$L_r = \frac{w\,n_1\,e_d}{\Delta_{max}}\,b_w = \frac{7.0(0.033)}{0.005}(0.75) = 34.7\ \text{m}\ \ (\text{Curve 1})$$ and 38.9 m for Curve 2. Both are shorter than the minimum runoff length, which TAC sets at two seconds of travel: $2(80/3.6) = 44.4$ m. The two-second criterion governs, and it is confirmed by the table's own spiral parameters, since $A^{2}/R$ gives $150^{2}/509.296 = 44.2$ m and $135^{2}/408.105 = 44.7$ m. Adopt $$\boxed{L_r = 45\ \text{m for both curves}}$$
Tangent runout length. The runout removes the normal crown at the same relative gradient, so it scales the runoff by the ratio of cross slopes: $$L_t = L_r\,\frac{e_{NC}}{e_d} \ \Rightarrow\ L_{t1} = 45\left(\tfrac{0.020}{0.033}\right) = 27.273\ \text{m}, \qquad L_{t2} = 45\left(\tfrac{0.020}{0.037}\right) = 24.324\ \text{m}$$
Distribution of the runoff about the PC. With no spiral the runoff straddles the tangent-to-curve point. AASHTO recommends placing 0.70 of the runoff on the tangent and 0.30 on the curve when two lanes are rotated, so $0.70(45) = 31.5$ m precedes the PC and $0.30(45) = 13.5$ m follows it. The transition points for Curve 1 are therefore $$\text{RC} = 120.000 - 31.5 = 0+088.500, \qquad \text{NC} = 88.500 - 27.273 = 0+061.227$$ $$\text{FS begins} = 120.000 + 13.5 = 0+133.500, \qquad \text{FS ends} = 520.000 - 13.5 = 0+506.500$$ and, mirrored beyond the PT, RC at 0+551.500 and NC at 0+578.773.
The same construction on Curve 2. Using $L_{t2} = 24.324$ m, $$\text{NC} = 0+584.176, \quad \text{RC} = 0+608.500, \quad \text{FS begins} = 0+653.500$$ $$\text{FS ends} = 0+947.025, \quad \text{RC} = 0+992.025, \quad \text{NC} = 1+016.349$$
Cross slope carried at each control point. Between RC and full superelevation the outside half rotates upward while the inside half holds the normal crown until the whole section is a single plane, which happens when the outside half reaches the normal 2.0 % — at $0.020/0.033 = 0.606$ of the runoff on Curve 1 and $0.020/0.037 = 0.541$ on Curve 2. Both fractions are smaller than the 0.70 that precedes the PC, so at each PC the pavement is already a plane carrying $0.70\,e$: $$\boxed{\text{PC}_1:\ 2.31\ \%\ \text{both halves}}\qquad \boxed{\text{PC}_2:\ 2.59\ \%\ \text{both halves}}$$ At NC the section is normally crowned at 2.0 % each way; at RC the outside half is level and the inside half still carries 2.0 %; between FS points the whole section lies at the full $e$.
Check that the transitions fit the available tangent. Leaving Curve 1 requires $31.5 + 27.273 = 58.77$ m and entering Curve 2 requires $31.5 + 24.324 = 55.82$ m, a total of 114.59 m against the 120.000 m of tangent available. The design fits with 5.40 m of full normal crown between the two runouts, which is the minimum acceptable condition when superelevation reverses sense.

Question 2 — superelevation transition schedule (rotation about the centreline, $L_r = 45$ m)
Point	Curve 1 station	Cross slope, Curve 1	Curve 2 station	Cross slope, Curve 2
NC (entering)	0+061.227	normal crown, 2.0 % each way	0+584.176	normal crown, 2.0 % each way
RC (entering)	0+088.500	outside 0.0 %, inside 2.0 %	0+608.500	outside 0.0 %, inside 2.0 %
PC	0+120.000	plane at 2.31 %	0+640.000	plane at 2.59 %
FS begins	0+133.500	full $e = 3.3$ %	0+653.500	full $e = 3.7$ %
FS ends	0+506.500	full $e = 3.3$ %	0+947.025	full $e = 3.7$ %
PT	0+520.000	plane at 2.31 %	0+960.525	plane at 2.59 %
RC (leaving)	0+551.500	outside 0.0 %, inside 2.0 %	0+992.025	outside 0.0 %, inside 2.0 %
NC (leaving)	0+578.773	normal crown restored	1+016.349	normal crown restored
Design superelevation	$e_1 = 0.033$ (R = 509.296 m)		$e_2 = 0.037$ (R = 408.105 m)
Tangent runout	27.273 m		24.324 m

Superelevation diagram for Curve 1. The solid line is the outside half of the pavement and the dashed line the inside half; the two coincide once the section becomes a single plane, which occurs before the PC.

Check: the runoff split and the relative gradient. AASHTO gives 0.70 as the recommended proportion of runoff placed on the tangent when two lanes are rotated, within an acceptable range of 0.6 to 0.9, and TAC accepts 0.6 to 0.8. Adopting the two-thirds value instead of 0.70 moves every NC and RC station by 1.5 m and leaves the cross slopes at the PC essentially unchanged (2.20 % and 2.47 %). The maximum relative gradient of 0.50 % at 80 km/h is likewise an AASHTO table value; because the two-second minimum governs here, the runoff length is insensitive to it.

Topic. Superelevation attainment on a simple curve without a spiral, on a pavement rotated about its centreline.

Key relations. $T = R\tan(\Delta/2)$; $L_c = R\Delta$ (radians); runoff $L_r = w n_1 e_d b_w/\Delta_{max}$ with $b_w = [1+0.5(n_1-1)]/n_1$; minimum runoff = two seconds of travel; runout $L_t = L_r\,e_{NC}/e_d$; runoff distribution 0.7 before / 0.3 after the PC.

Why this works. Superelevation cannot appear instantaneously: the outside edge of pavement has to be raised relative to the centreline, and the rate at which the edge profile may depart from the centreline profile is limited by appearance and by drainage, which is what the maximum relative gradient expresses. Rotating about the centreline halves the edge movement compared with rotating about an edge, and the $b_w$ factor recognises that the outer lanes of a wide pavement need not each be given the full gradient. Placing most of the runoff on the tangent means the driver reaches the PC with most of the superelevation already developed, which matches where the lateral acceleration actually builds up as the steering input is applied.

Common pitfall. Forgetting to check the two-second minimum, which governs whenever the design superelevation is modest, and so reporting a runoff of 35 m instead of 45 m. The second pitfall is to report a single cross slope at RC: the outside half is level there but the inside half still carries the full normal crown, and saying so is what earns the mark.

Source. TAC, Geometric Design Guide for Canadian Roads, §2.1.2 (Table 2.1.2.5, page 2.1.2.12); AASHTO Green Book, §3.3 (Tables 3-15 to 3-18); Garber & Hoel, §15.4.

← Question 1 Next: Question 3 →