16-Civ-B7 Transportation Planning and Engineering · December 2013

Question 1 of 7

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 2013 — a three-hour open-book examination; any non-communicating calculator is permitted. The cover page states that a total of five solutions is required, that only the first five as they appear in the answer book will be marked, and that all questions are of equal value. The grading scheme printed on page 1 confirms 20 marks per question, split as: Q1 20; Q2 20; Q3 (a) 15 and (b) 5; Q4 (a) 8 and (b) 12; Q5 (a) 8 and (b) 12; Q6 20; Q7 (a) 8 and (b) 12. All seven printed questions are worked below, because this set is a study resource rather than a timed attempt; on exam day a candidate submits only the first five, in order. The paper also states that any data required but not given may be assumed and that assumptions should be recorded with the answer — several questions need that licence, and every assumption is flagged where it is made.

Reference texts. N.J. Garber and L.A. Hoel, Traffic and Highway Engineering, 5th ed. (sight distance, vertical and horizontal alignment, traffic stream models, earthwork); Transportation Association of Canada, Geometric Design Guide for Canadian Roads (Canadian design-domain values for stopping sight distance, perception-reaction time and deceleration); AASHTO, A Policy on Geometric Design of Highways and Streets (the tabulated metric stopping sight distances); AASHTO, Guide for Design of Pavement Structures (1993) (rigid pavement thickness, reliability, drainage and load-transfer coefficients); Asphalt Institute, Mix Design Methods MS-2 (gradation charts, the 0.45 power chart, aggregate blending); M.S. Mamlouk and J.P. Zaniewski, Materials for Civil and Construction Engineers, 4th ed. (aggregate moisture states, sieve analysis); Transportation Association of Canada, Pavement Asset Design and Management Guide (Canadian pavement design practice).

Check — assumptions carried through this paper. Four inputs the exam does not supply are assumed under its own Note 2 (“any data required, but not given, can be assumed”), and each is restated at the point of use: (i) Question 2 needs a stopping-sight-distance basis — a 2.5 s perception-reaction time and a 3.4 m/s² deceleration, the TAC and AASHTO design values, giving the tabulated 185 m at 100 km/h; (ii) Question 2 also needs to know whether the 600 m radius is to the road centreline — it is taken as the centreline, and Step 5 shows the alternative reading changes the answer by 0.02 m; (iii) Question 6 does not say whether the transverse joints are dowelled — dowels are assumed, giving a load-transfer coefficient J = 3.2, with the undowelled case quantified in a callout; (iv) Question 6 gives a drainage description rather than a coefficient, so C_d = 1.00 is read from the AASHTO table, again with the alternative quantified.

Question 1 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. An equal-tangent sag parabola is defined by its length, the station and elevation of its beginning, and the two tangent grades:

Given data — sag vertical curve
Quantity	Symbol	Value
Curve length	$L$	300.000 m
Station of PVC	—	2+600.000
Elevation of PVC	$Y_{PVC}$	320.000 m
Approach (back) grade	$g_1$</td><td>$-4.0\%$
Departure (forward) grade	$g_2$</td><td>$+1.0\%$

Find. The station and elevation of the PVI, of the PVT and of the lowest point on the curve, and the station and elevation of the four points at which the grade of the curve is $-3\%$, $-2\%$, $-1\%$ and $0\%$.

Sag vertical curve: PVC, PVI, PVT, the four requested grade points and the lowest point at station 2+840.000.

Approach. Treat the curve as an equal-tangent parabola so that the grade varies linearly from $g_1$ to $g_2$ over the length $L$; every station asked for is then found by locating the distance $x$ from the PVC at which the required grade occurs, and substituting that $x$ into the single parabolic elevation equation.

Establish the algebraic difference in grade and the rate at which grade changes. For a vertical curve the algebraic difference is taken with signs attached, so $$A = g_2 - g_1 = (+1.0) - (-4.0) = +5.0\%$$ The grade therefore changes at the constant rate $$r = \frac{A}{L} = \frac{5.0\%}{300\text{ m}} = 0.016667\ \%/\text{m}$$ and the curve is a sag curve because $A$ is positive. The rate-of-change parameter $K = L/A = 300/5.0 = 60$ m per percent.
Locate the PVI, which on an equal-tangent curve sits half the curve length ahead of the PVC. $$\text{Sta}_{PVI} = \text{Sta}_{PVC} + \frac{L}{2} = (2+600.000) + 150.000 = \boxed{2+750.000}$$ Its elevation follows the back tangent down from the PVC: $$Y_{PVI} = Y_{PVC} + \frac{g_1}{100}\left(\frac{L}{2}\right) = 320.000 + (-0.0400)(150.000) = 314.000\text{ m}$$
Locate the PVT by continuing along the forward tangent from the PVI. $$\text{Sta}_{PVT} = \text{Sta}_{PVC} + L = (2+600.000) + 300.000 = \boxed{2+900.000}$$ $$Y_{PVT} = Y_{PVI} + \frac{g_2}{100}\left(\frac{L}{2}\right) = 314.000 + (+0.0100)(150.000) = 315.500\text{ m}$$
Write the elevation of the curve as a single function of the distance $x$ measured from the PVC. The parabola is the back tangent plus a second-order offset that grows as $A x^2 / 200L$: $$Y(x) = Y_{PVC} + \frac{g_1}{100}x + \frac{A}{200L}x^2$$ $$Y(x) = 320.000 - 0.0400\,x + \frac{5.0}{200(300)}x^2 = 320.000 - 0.0400\,x + \frac{x^2}{12\,000}$$ As a check, $Y(300) = 320.000 - 12.000 + 7.500 = 315.500$ m, which reproduces the PVT elevation obtained in Step 3.
Find the lowest point, where the tangent to the curve is horizontal. Differentiating the elevation equation, or equivalently setting the running grade $g(x) = g_1 + (A/L)x$ to zero, $$x_{low} = \frac{-g_1 L}{A} = \frac{(4.0)(300)}{5.0} = 240.000\text{ m}$$ $$\text{Sta}_{low} = (2+600.000) + 240.000 = \boxed{2+840.000}$$ $$Y_{low} = 320.000 - 0.0400(240.000) + \frac{240.000^2}{12\,000} = 320.000 - 9.600 + 4.800 = 315.200\text{ m}$$
Locate the four requested grade points by inverting the linear grade equation. Setting $g(x) = g_1 + (A/L)x$ equal to each required grade gives $$x = \frac{(g - g_1)L}{A}$$ so that $x = 60.000$ m at $-3\%$, $120.000$ m at $-2\%$, $180.000$ m at $-1\%$ and $240.000$ m at $0\%$ — the grade points are evenly spaced at 60 m because the grade changes linearly. Substituting each $x$ into the elevation equation of Step 4 gives the elevations tabulated below; for example, at $-2\%$, $$Y(120.000) = 320.000 - 0.0400(120.000) + \frac{120.000^2}{12\,000} = 320.000 - 4.800 + 1.200 = 316.400\text{ m}$$
Check the results against the symmetry of the parabola. The curve is symmetric about its lowest point at $x = 240$ m, so stations equidistant from it must share an elevation. The $-1\%$ point at $x = 180$ m and the PVT at $x = 300$ m are each 60 m from the low point, and both come out at 315.500 m, confirming the arithmetic. The $0\%$ grade point is, as expected, the lowest point itself.

Two design checks are worth adding even though the question does not ask for them, because an examiner rewards the candidate who notices that the curve has to work as well as compute. For a sag curve at 100 km/h the headlight criterion requires $L = AS^2/(120 + 3.5S)$; with $S = 185$ m and $A = 5.0$ this gives 223 m, and the comfort criterion $L = AV^2/(100a)$ with $a = 0.3$ m/s² gives 167 m. The 300 m curve provided comfortably satisfies both, so the geometry is not merely arithmetically consistent but also adequate for the design speed.

Final results — Question 1
Point	Grade at the point	Station	Elevation (m)
PVC (given)	$-4.0\%$	2+600.000	320.000
Grade point	$-3.0\%$	2+660.000	317.900
Grade point	$-2.0\%$	2+720.000	316.400
PVI	tangent intersection	2+750.000	314.000
Grade point	$-1.0\%$	2+780.000	315.500
Lowest point ($0\%$ grade)</td><td>$0.0\%$	2+840.000	315.200
PVT	$+1.0\%$	2+900.000	315.500

Topic. Equal-tangent parabolic vertical curves: stationing, elevations, running grade and the location of the high or low point.

Key relations. $A = g_2 - g_1$ (signed); $Y(x) = Y_{PVC} + (g_1/100)x + (A/200L)x^2$; running grade $g(x) = g_1 + (A/L)x$; turning point $x = -g_1L/A$; rate-of-change parameter $K = L/A$; and for the PVI, $\text{Sta}_{PVI} = \text{Sta}_{PVC} + L/2$ on an equal-tangent curve only.

Why this works. A parabola is chosen for vertical curves precisely because its first derivative — the grade — is linear in distance, so the rate of change of grade is constant and a vehicle experiences a constant vertical acceleration through the curve. That single property is what makes every result above a one-line substitution: the grade points are equally spaced, the turning point is found by setting a linear function to zero, and elevations come from one quadratic. The offset term $A x^2/200L$ carries the factor 200 rather than 100 because $A$ is expressed in percent and the term integrates the linear grade change over $x$.

Common pitfall. Taking $A$ as $|g_2 - g_1|$ or as $g_1 - g_2$ inverts the sign of the offset and turns a sag curve into a crest. A second trap is assuming the lowest point falls at mid-curve; it does so only when $g_1 = -g_2$. Here it lies at $x = 240$ m, well past the 150 m midpoint, because the back grade is four times steeper than the forward grade. Third, in metric stationing the number after the plus sign is metres, not 20-m or 100-ft units, so $2+600.000 + 240.000$ is $2+840.000$ and not $4+... $ — carry the metres straight through.

Source. N.J. Garber and L.A. Hoel, Traffic and Highway Engineering, 5th ed., Chapter 3 (Geometric Design of Highway Facilities), vertical alignment; Transportation Association of Canada, Geometric Design Guide for Canadian Roads, sag and crest curve design domains.

← Paper overview Next: Question 2 →