16-Civ-B7 Transportation Planning and Engineering · December 2013
Question 2 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 Cd = 1.00 is read from the AASHTO
table, again with the alternative quantified.
Given. A simple circular curve on a two-lane rural highway, with
the radius quoted to the road centreline:
Given data — horizontal curve sight line
Quantity
Symbol
Value
Radius to the road centreline
$R$
600 m
Lane width
$w$
3.5 m
Number of lanes
—
2 (one each way)
Design speed
$V$
100 km/h
Perception-reaction time (assumed)
$t$
2.5 s
Deceleration rate (assumed)
$a$
3.4 m/s2
Find. The lateral distance, measured from the inside edge of the
inside lane, over which sight obstructions must be removed so that a driver in the
inside lane always has the full stopping sight distance available.
Sight line across the inside of the horizontal curve. The clearance is measured from the inside edge of the inside lane, not from the vehicle path.
Approach. Compute the stopping sight distance for 100 km/h, treat
that distance as an arc travelled along the driver’s own path (the centre of the
inside lane), find the middle ordinate of the corresponding chord, and then convert
that ordinate — which is measured from the vehicle path — into a clearance
measured from the inside edge of the lane by deducting half a lane width.
Establish the stopping sight distance for the design speed. The
standard two-term expression adds the distance covered during perception-reaction to
the braking distance:
$$S = 0.278\,Vt + \frac{V^2}{254\,(a/9.81)}$$
$$S = 0.278(100)(2.5) + \frac{100^2}{254\,(3.4/9.81)} = 69.5 + 113.6 = 183.1\text{ m}$$
Both TAC and AASHTO round this up to the tabulated design value, so
$$S = \boxed{185\text{ m}}$$
is carried forward. (Using the computed 183.1 m instead changes the final clearance by
about 0.11 m, so the choice is not critical.)
Identify the radius of the path the driver actually follows.
Sight distance is measured along the centre of the inside lane, not along the road
centreline. Because the inside lane lies half a lane width inside the centreline,
$$R_v = R - \frac{w}{2} = 600 - \frac{3.5}{2} = 598.25\text{ m}$$
Compute the middle ordinate of the sight-line chord. The line of
sight is the chord joining the driver and the object, both on the vehicle path; the
sight obstruction must be cleared back to that chord. For an arc of length $S$ on a
circle of radius $R_v$, the half angle subtended is $28.65\,S/R_v$ degrees, and
$$M = R_v\left[1 - \cos\!\left(\frac{28.65\,S}{R_v}\right)\right]$$
$$M = 598.25\left[1 - \cos\!\left(\frac{28.65 \times 185}{598.25}\right)\right]
= 598.25\,\bigl[1 - \cos(8.8596^\circ)\bigr]$$
$$M = 598.25\,(0.0119345) = \boxed{7.14\text{ m}}$$
The constant 28.65 is simply $90/\pi$, which converts the half-arc $S/2R_v$ from
radians into degrees.
Convert the middle ordinate into a clearance measured from the lane
edge. $M$ is measured inward from the vehicle path, which is itself
$w/2 = 1.75$ m inside the lane’s inner edge, so the width that must actually be
cleared beyond the pavement is
$$m = M - \frac{w}{2} = 7.14 - 1.75 = \boxed{5.39\text{ m}}$$
Rounding up for construction, the clear-sight (daylight) zone should extend
5.4 m from the inside edge of the inside lane, and nothing higher than
the 0.6 m object height — barriers, sign supports, guide rail, vegetation, cut
slopes or noise walls — may stand inside it.
Test the assumption made about the radius. The question does not
say whether the 600 m applies to the centreline or to the inside lane. If instead the
600 m is already the vehicle-path radius, then
$M = 600\,[1 - \cos(28.65 \times 185/600)] = 7.12$ m and the clearance becomes 5.37 m,
a difference of only 0.02 m. The answer is therefore insensitive to that ambiguity,
and 5.4 m is safe under either reading.
Confirm that the geometry the formula assumes is the geometry that
exists. The middle-ordinate expression assumes the whole sight line lies
within the circular curve, which requires the curve to be at least $S = 185$ m long.
At 600 m radius that is a central angle of $185/600$ radians, about
$17.7^\circ$; any curve with a smaller deflection would place the driver or the object
on the tangent, where the required clearance is less and a graphical or two-part
solution is needed. The 5.39 m result should therefore be quoted together with the
condition that the curve subtends at least about $18^\circ$.
Check — values assumed, not given. The exam supplies only
speed, radius and lane width, so the perception-reaction time (2.5 s) and the
deceleration rate (3.4 m/s2) are taken from TAC and AASHTO practice for
rural highways under the paper’s Note 2. A 0.6 m object height and a 1.08 m eye
height are implied by the same standards; they do not enter the arithmetic but they do
define what counts as an obstruction inside the 5.4 m zone. If a provincial standard
imposed a longer sight distance — say 210 m for a wet-weather or truck-braking
criterion — the clearance would grow roughly with $S^2$, to about 7.5 m.
Final results — Question 2
Quantity
Symbol
Value
Stopping sight distance at 100 km/h
$S$
185 m (183.1 m computed)
Radius of the vehicle path (centre of inside lane)
$R_v$
598.25 m
Half angle subtended by the sight chord
—
$8.8596^\circ$
Middle ordinate from the vehicle path
$M$
7.14 m
Clearance required from the inside edge of the inside lane