16-Civ-A6 Highway Design, Construction, and Maintenance · May 2017

Question 2 of 7: Horizontal curve on a two-lane highway — radius check, restricted clearance and stationing

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

Notes on this paper

Paper format. National Examinations, May 2017 — 16-Civ-A6 Highway Design, Construction and Maintenance. Three-hour closed-book paper (Casio or Sharp approved calculator only). Seven questions, all of equal value at 20 marks; the candidate submits five, and only the first five in the answer book are marked. NOTE 1 invites a written statement of any assumption, and NOTE 2 permits any datum that is required but not given to be assumed. All seven questions are worked below, because the set is a study resource rather than an examination script.

Reference texts.

Garber, N. J. and Hoel, L. A., Traffic and Highway Engineering, 5th ed., Cengage — the exam's equation sheet, the friction/side-friction table, Table 4.2 and Table 4.4 axle-load equivalency factors, Table 4.5 reliability Z_R values and the structural-layer-coefficient table are all reproduced from this text.
AASHTO, Guide for Design of Pavement Structures, Washington DC, 1993 — the flexible-pavement design equation and nomograph, and the source of the LEF tables.
AASHTO, A Policy on Geometric Design of Highways and Streets (the “Green Book”) — stopping sight distance, horizontal sightline offset and vertical-curve K values.
Transportation Association of Canada, Geometric Design Guide for Canadian Roads (TAC GDG) — the governing Canadian geometric-design reference; its design-domain values parallel the AASHTO relations used here.
Transportation Association of Canada, Manual of Uniform Traffic Control Devices for Canada (MUTCDC) — advance placement of guide signs.

Canadian context. These are Engineers Canada national examinations, so the Canadian frame governs practice: geometric design in British Columbia follows the TAC Geometric Design Guide and signing follows the MUTCDC. The paper, however, supplies an AASHTO equation sheet and AASHTO tables and names the AASHTO method explicitly in Questions 4 and 5, so every calculation below is carried out with the data the paper gives; where a Canadian practice differs in emphasis it is noted in the concept block rather than substituted for the examiner's method.

Assumptions used throughout (declared under NOTE 1). Gravitational acceleration $g = 9.81\ \text{m/s}^{2}$; perception–reaction time $t_{pr} = 2.5\ \text{s}$ (the AASHTO design value — the paper does not state one); coefficients $f$ and $f_{s}$ read from the paper's own Table 1 at the initial vehicle speed; drainage coefficients $m_{2} = m_{3} = 1.0$ where the question is silent.

Question 2: Horizontal curve on a two-lane highway — radius check, restricted clearance and stationing (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 two-lane two-way rural curve on a constant grade, with the superelevation capped by local policy and the radius already proposed by a preliminary design.

Given data
Item	Symbol	Value
Design speed	$V$	$80\ \text{km/h} = 22.222\ \text{m/s}$
Proposed radius (centreline)	$R$	$240\ \text{m}$
Maximum superelevation	$e_{\max}$	$0.08$
Side-friction factor (Table 1, 80 km/h)	$f_{s}$	$0.14$
Central angle	$\Delta$	$30^\circ$
Lane width	$w$	$3.6\ \text{m}$
Grade on the inside (up-grade) lane	$G$	$+2\ \%$
Clearable offset from the centreline	—	$6\ \text{m}$
Stationing of the PI	—	$100+00$

Find. Whether $R = 240\ \text{m}$ is adequate at 80 km/h, the speed limit that the restricted 6 m excavation actually supports, and the station of the PT.

Plan of the curve. The excavation limit lies 6 m inside the centreline; the sight line is measured from the inside-lane vehicle path, so the usable middle ordinate is $6.0-1.8 = 4.2\ \text{m}$.

Approach. Compare the proposed radius against the minimum that lateral equilibrium allows at the design speed; then invert the horizontal-sightline-offset relation to find what sight distance the restricted excavation leaves, and read back through the stopping-sight-distance expression to the highest speed that distance supports; finally use the tangent length and curve length to chain the stationing forward from the PI.

Test the proposed radius against the design speed. With $e = e_{\max} = 0.08$ and $f_{s} = 0.14$, $$R_{\min}=\frac{V^{2}}{g\,(f_{s}+e)}=\frac{22.222^{2}}{9.81\,(0.14+0.08)}=\frac{493.83}{2.1582}=\boxed{228.81\ \text{m}}$$ Because $240\ \text{m} \gt 228.81\ \text{m}$, the preliminary radius is adequate.
Express the same check as a friction demand. It is worth reporting the result the other way round, because it shows how much reserve there is: at $R = 240\ \text{m}$ the curve demands $$f_{s,\text{req}}=\frac{V^{2}}{gR}-e=\frac{493.83}{9.81(240)}-0.08=0.2098-0.08=0.1297$$ against the $0.14$ the table allows, so about 7 % of the side-friction allowance is held in reserve. The radius is safe, but only just, which is why part (b) matters.
Find the sight distance the restricted excavation leaves. The driver whose view is critical travels in the inside lane, whose centreline lies half a lane inside the road centreline, so $$R_{v}=240-1.8=238.2\ \text{m},\qquad M_{s}=6.0-1.8=4.2\ \text{m}$$ Inverting the middle-ordinate relation, $$S=\frac{\pi R_{v}}{90}\,\cos^{-1}\!\left(1-\frac{M_{s}}{R_{v}}\right)=\frac{\pi(238.2)}{90}\,\cos^{-1}(0.98237)$$ The inverse cosine is $10.765^\circ$, so $$\boxed{S=89.59\ \text{m}}$$ of sight distance is available — far less than the 80 km/h requirement.
Read back to a speed. The available distance must cover stopping from the posted speed on the $+2\ \%$ inside lane. Testing the tabulated speeds, $$\text{SSD}_{60}=\frac{16.667^{2}}{2(9.81)(0.33+0.02)}+16.667(2.5)=40.45+41.67=82.12\ \text{m}$$ $$\text{SSD}_{70}=\frac{19.444^{2}}{2(9.81)(0.31+0.02)}+19.444(2.5)=58.40+48.61=107.01\ \text{m}$$ Since $82.12 \lt 89.59 \lt 107.01$, the site supports 60 km/h but not 70 km/h, so $$\boxed{\text{post a 60 km/h speed limit through the curve}}$$
Compute the tangent and curve lengths. $$T=R\tan\frac{\Delta}{2}=240\tan 15^\circ=64.31\ \text{m}$$ $$L=\frac{\pi\Delta R}{180}=\frac{\pi(30)(240)}{180}=125.66\ \text{m}$$
Chain the stationing through the curve. The PC lies one tangent length back from the PI and the PT one curve length forward of the PC (not one tangent length forward of the PI — stationing follows the alignment, which runs along the arc): $$\text{Sta}_{PC}=10\,000.00-64.31=9\,935.69\ \text{m}\;(99+35.69)$$ $$\text{Sta}_{PT}=9\,935.69+125.66=10\,061.36\ \text{m}$$ $$\boxed{\text{PT}=100+61.36}$$

Posting 60 km/h is an operational remedy, not the only one. If the 80 km/h design speed must be preserved, the required sight distance is $139.45\ \text{m}$, which needs a middle ordinate of $10.14\ \text{m}$ from the inside-lane path, that is $11.94\ \text{m}$ from the centreline — nearly double the 6 m the budget allows. Presenting both the posted-speed answer and the excavation the alternative would cost is what turns the arithmetic into a design recommendation.

Final results
Quantity	Symbol	Value
Minimum radius at 80 km/h	$R_{\min}$	$228.81\ \text{m}$ (240 m is adequate)
Side friction demanded at $R=240$ m	$f_{s,\text{req}}$	$0.1297 \lt 0.14$
Sight distance with 6 m cleared	$S$	$89.59\ \text{m}$
SSD required at 60 / 70 km/h on $+2\%$	$\text{SSD}$	$82.12$ / $107.01\ \text{m}$
Speed limit to be posted	—	$60\ \text{km/h}$
Tangent length	$T$	$64.31\ \text{m}$
Curve length	$L$	$125.66\ \text{m}$
Station of the PC	—	$99+35.69$
Station of the PT	—	$100+61.36$

Check: the grade sign matters. The question states that the inside lane is up-grade, and the inside lane is the one whose sight line is obstructed by the excavation, so $G=+0.02$ is used. Taking $-2\ \%$ instead would give $\text{SSD}_{70}=88.4\ \text{m}$, which sneaks under the available $89.59\ \text{m}$ and would lead to posting 70 km/h — an unsafe answer produced by a sign error. Also assumed: $t_{pr}=2.5\ \text{s}$, and the obstruction is continuous through the curve, so the middle ordinate rather than a single point offset governs.

Topic. Checking a preliminary curve radius against the design speed, and converting a construction constraint into an operating-speed restriction.

Key relations. $R_{\min}=V^{2}/[g(f_{s}+e)]$; $f_{s,\text{req}}=V^{2}/(gR)-e$; $M_{s}=R_{v}\{1-\cos[90S/(\pi R_{v})]\}$ and its inverse $S=(\pi R_{v}/90)\cos^{-1}(1-M_{s}/R_{v})$; $T=R\tan(\Delta/2)$; $L=\pi\Delta R/180$; $\text{Sta}_{PT}=\text{Sta}_{PI}-T+L$.

Why this works. A horizontal curve has to satisfy two independent controls. The first is dynamic: the combination of superelevation and side friction must hold the vehicle on its path, and this sets a minimum radius. The second is visual: the driver must be able to see far enough along the inside of the curve to stop, and this sets a minimum clear offset. A radius that passes the first check can fail the second, which is precisely the situation here, and the remedy is either more excavation or a lower posted speed. The stationing arithmetic differs from the tangent geometry because chainage is measured along the alignment: the curve replaces the two tangents, so the PT is $L$ beyond the PC, and $L \lt 2T$ always.

Common pitfall. Adding the tangent length to the PI station to get the PT is the classic slip; it gives $100+64.31$ instead of $100+61.36$, and the difference ($2T-L = 3.0\ \text{m}$ here) grows quickly with the central angle. The other frequent error is measuring both the radius and the clearance from the road centreline; the sight line belongs to the vehicle in the inside lane, so half a lane must be taken off both.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §15.7 (simple curves and stationing) and §15.5 (sight distance on horizontal curves); TAC Geometric Design Guide for Canadian Roads, Ch. 3.

← Question 1 Next: Question 3 →