16-Civ-A6 Highway Design, Construction, and Maintenance · May 2017
Question 1 of 7: Interchange ramp — minimum radius, lateral 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 ZR 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.
Given. A two-lane connector ramp turning through a right angle between an overpass at A and the lower freeway at B, designed for a low ramp speed on full superelevation.
Given data
Item
Symbol
Value
Design speed
$V$
$30\ \text{km/h} = 8.333\ \text{m/s}$
Superelevation
$e$
$0.08$
Side-friction factor (Table 1, 30 km/h)
$f_{s}$
$0.17$
Longitudinal friction factor (Table 1, 30 km/h)
$f$
$0.40$
Lane width, two lanes
$w$
$3.6\ \text{m}$ each
Central angle (from the figure)
$\Delta$
$90^\circ$
Elevation drop, A to B
$\Delta z$
$5\ \text{m}$
Perception–reaction time (assumed)
$t_{pr}$
$2.5\ \text{s}$
Stationing of the PT
—
$90+00$
Find. The minimum centreline radius, the lateral clearance that must be provided beyond the inside edge of the ramp so that stopping sight distance is available around the curve, and the station of the PC.
Plan of the ramp. The dashed blue arc is the centreline of the two lanes, on which the design radius is measured; the green chord is the sight line whose middle ordinate $M_{s}$ sets the clearance.
Approach. Balance the lateral forces on a superelevated curve to size the radius, convert that radius into a curve length so the 5 m elevation drop becomes a ramp grade, use the grade in the stopping-sight-distance expression, and then convert the required sight distance into a horizontal sightline offset that is finally referred to the inside edge of the pavement.
Convert the design speed to SI. The equation sheet works in metres and seconds, so $$V=\frac{30}{3.6}=8.333\ \text{m/s}$$
Size the curve from lateral equilibrium. On a superelevated curve the side friction and the superelevation together supply the centripetal acceleration, giving the sheet's relation $$R_{v}=\frac{V^{2}}{g\,(f_{s}+e)}=\frac{8.333^{2}}{9.81\,(0.17+0.08)}=\frac{69.44}{2.4525}$$ so that $$\boxed{R_{\min}=28.32\ \text{m}}$$ measured, as the question directs, on the centreline of the two lanes. This is a tight loop ramp, which is exactly what a 30 km/h design speed buys.
Convert the radius into a curve length. The figure shows the ramp turning through a right angle, so $$L=\frac{\pi\,\Delta\,R}{180}=\frac{\pi\,(90)(28.32)}{180}=44.48\ \text{m}$$
Turn the elevation difference into a grade. The 5 m drop from the overpass at A to the freeway at B is taken over that curve, the only length the question defines, so the ramp descends at $$G=-\frac{5}{44.48}=-0.1124\quad(-11.24\ \%)$$ A downgrade lengthens the braking distance and must not be dropped.
Compute the stopping sight distance on that grade. Using the sheet's expression with $f=0.40$ at 30 km/h, $$\text{SSD}=\frac{V_{0}^{2}}{2g\,(f+G)}+V_{0}t_{pr}=\frac{69.44}{2(9.81)(0.40-0.1124)}+8.333(2.5)$$ $$\text{SSD}=12.31+20.83=\boxed{33.14\ \text{m}}$$ Note how little of this is braking: at 30 km/h the reaction distance is nearly two-thirds of the total.
Convert the sight distance into a horizontal sightline offset. The driver sights across the inside of the curve along a chord, and the largest gap between that chord and the travelled path is the middle ordinate $$M_{s}=R_{v}-R_{v}\cos\!\left[\frac{90\,\text{SSD}}{\pi R_{v}}\right]=28.32\left[1-\cos\left(\frac{90(33.14)}{\pi(28.32)}\right)\right]$$ The bracket evaluates to $33.53^\circ$, and $\cos 33.53^\circ = 0.8337$, so $$M_{s}=28.32\,(1-0.8337)=4.71\ \text{m}$$ measured from the ramp centreline.
Refer the offset to the inside edge of the pavement. The centreline of the two lanes sits one full lane, $3.6\ \text{m}$, inside of the inside edge, so the strip that must actually be cleared of barriers, cut slopes, noise walls and vegetation beyond the pavement edge is $$\Delta M = M_{s}-w = 4.71-3.60 = \boxed{1.11\ \text{m}}$$
Chain the stationing back from the PT. Stations run forward along the alignment, so the PC lies one curve length upstream of the PT: $$\text{Sta}_{PC}=\text{Sta}_{PT}-L = 9\,000.00-44.48 = 8\,955.52\ \text{m}$$ $$\boxed{\text{PC} = 89+55.52}$$
The three answers hang together: the tight radius forced by the 30 km/h design speed is what makes the curve short, and the short curve is what makes the ramp steep, which in turn is what stretches the sight distance and therefore the clearance width.
Final results
Quantity
Symbol
Value
Minimum centreline radius
$R_{\min}$
$28.32\ \text{m}$
Length of circular curve
$L$
$44.48\ \text{m}$
Ramp grade, A to B
$G$
$-11.24\ \%$
Stopping sight distance on the grade
$\text{SSD}$
$33.14\ \text{m}$
Horizontal sightline offset from the centreline
$M_{s}$
$4.71\ \text{m}$
Width to be cleared beyond the inside edge
$\Delta M$
$1.11\ \text{m}$
Station of the PC
—
$89+55.52$
Check: three declared assumptions. (i) $t_{pr}=2.5\ \text{s}$ is the AASHTO design value; the paper gives none, and NOTE 2 permits the assumption. (ii) The 5 m elevation difference is taken over the 44.48 m circular curve, because A and B are the ends of that curve and no other length is dimensioned; if the drop were spread over a longer ramp including tangents the grade would be flatter and the required clearance smaller, so this reading is conservative. (iii) The middle ordinate has been computed on the radius the question defines, the centreline of the two lanes. If instead the sight line is taken along the inside-lane centreline ($R_{v}=28.32-1.80=26.52\ \text{m}$), then $M_{s}=5.01\ \text{m}$ and the clearance beyond the inside edge becomes $5.01-1.80=3.21\ \text{m}$. Either reading is defensible on this wording; the figure and the results table quote the first.