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

Question 3 of 8: Moving the Low Point of a Sag Curve onto a Catch Basin

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 3: Moving the Low Point of a Sag Curve onto a Catch Basin (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 existing symmetrical sag curve of length $L = 80\ \text{m}$ with $g_1 = -5\ \%$ and $g_2 = +8\ \%$; PVI at station 0+150 with elevation 26.50 m; a catch basin fixed at station 0+139 with a grate at elevation 27.74 m; the $-5\ \%$ grade line is fixed, the remainder of the geometry may change, and the replacement curve must be symmetrical.

Find. A change to the vertical geometry that puts the low point of the curve exactly at station 0+139, and the grate elevation the adjusted catch basin must be set to.

Existing 80 m sag curve (dashed) and the proposed 95.333 m curve (solid). Lengthening the curve while holding both grade lines and the PVI draws the low point back onto the catch basin and lifts the profile there by 233 mm.

Approach. Locate the low point of the existing curve to confirm the diagnosis, then impose the condition that the low point coincide with the catch basin. Because the $-5\ \%$ grade is fixed and the curve must stay symmetrical, the cleanest single change is to lengthen the curve while keeping both grade lines and the PVI in place; the new grate elevation then follows from the curve equation.

Set up the existing curve. The algebraic grade difference is $A = g_2 - g_1 = 8 - (-5) = 13\ \%$. For a symmetrical curve the PVC lies half a length back from the PVI: $$\text{PVC} = 150 - \tfrac{80}{2} = 0+110.000, \qquad \text{elev}_{PVC} = 26.50 + 0.05(40) = 28.500\ \text{m}$$ and the profile is the parabola $$y(x) = \text{elev}_{PVC} + g_1 x + \frac{A}{200L}x^{2} = 28.500 - 0.05x + 0.00081250\,x^{2}$$ with $x$ measured from the PVC.
Confirm the diagnosis. The low point is where the profile gradient vanishes, at $$x_{low} = \frac{L\,|g_1|}{A} = \frac{80(5)}{13} = 30.769\ \text{m} \ \Rightarrow\ \text{station } 0+140.769$$ The road surface at the catch basin ($x = 29.000$ m) is $y = 28.500 - 1.450 + 0.683 = 27.733$ m, while the low point sits 1.77 m downstream at elevation 27.731 m. The grate at 27.740 m is therefore 7 mm above the pavement beside it as well as being off the sag, so water both bypasses the inlet and ponds against it — exactly the reported flooding.
Impose the design condition. Holding the PVI at station 0+150 and both grades, the low point of a symmetrical curve of unknown length $L$ falls at $$\text{station}_{low} = \text{PVI} - \frac{L}{2} + \frac{L\,|g_1|}{A} = 150 - L\left(0.5 - \frac{5}{13}\right) = 150 - 0.115385\,L$$ Setting this equal to 139.000 gives $$L = \frac{11.000}{0.115385} = \boxed{95.333\ \text{m}}$$ so the curve is lengthened by 15.333 m and remains symmetrical, as required. This is the assumption adopted: retain both grade lines and the PVI, and lengthen the curve.
Re-establish the new curve. The new PVC moves back and up along the fixed $-5\ \%$ grade: $$\text{PVC} = 150 - \tfrac{95.333}{2} = 0+102.333, \qquad \text{elev}_{PVC} = 26.50 + 0.05(47.667) = 28.883\ \text{m}$$ and the PVT moves forward to station 0+197.667. The low point is now at $x = 95.333(5)/13 = 36.667$ m from the PVC, which is station 0+139.000 — the catch basin, as designed.
New grate elevation. Evaluating the new parabola at the low point, $$y = 28.883 - 0.05(36.667) + \frac{13}{200(95.333)}(36.667)^{2}$$ $$y = 28.883 - 1.833 + 0.917 = \boxed{27.967\ \text{m}}$$ The grate must therefore be raised from 27.740 m to 27.967 m, an adjustment of $+0.227$ m. Because the frame and grate sit on adjustment rings, a 227 mm lift is well within what a reconstruction can deliver without rebuilding the barrel.
Check the curve that results. The rate of vertical curvature is $K = L/A = 95.333/13 = 7.33$. Sag curves on a local street are usually governed by headlight sight distance, $K = S^{2}/(120 + 3.5S)$, which needs $K = 5.1$ at 30 km/h, 8.5 at 40 km/h and 12.2 at 50 km/h. The lengthened curve serves roughly 35 km/h, so it suits a residential street but should be checked against the posted speed before the profile is adopted; lengthening further only improves it.
Alternatives, for completeness. Two other single changes satisfy the same condition and are worth quoting because they cost far less lift. Holding $L = 80$ m and steepening the departure grade to $g_2 = +8.79\ \%$ moves the low point to 0+139 and gives a grate elevation of 27.775 m, a rise of only 35 mm, at the price of a steeper street beyond the sag. Holding $L = 80$ m and $g_2 = +8\ \%$ but sliding the PVI back along the fixed $-5\ \%$ grade to station 0+148.231 (elevation 26.588 m) gives 27.822 m, a rise of 82 mm. Lengthening the curve is nonetheless the preferred answer: it changes no grade the adjacent properties depend on, it improves the sag's comfort and headlight performance rather than degrading them, and it is the only one of the three that leaves the design better than it was.

Question 3 — final results
Quantity	Existing	Proposed
Curve length $L$	80.000 m	95.333 m
PVC station and elevation	0+110.000, 28.500 m	0+102.333, 28.883 m
PVT station	0+190.000	0+197.667
Low-point station	0+140.769	0+139.000 (at the catch basin)
Road elevation at station 0+139	27.733 m	27.967 m
Grate elevation	27.740 m (7 mm proud)	27.967 m (raise 0.227 m)
Rate of vertical curvature $K$	6.15	7.33
Alternative: steepen $g_2$ to 8.79 %	grate 27.775 m (raise 0.035 m)
Alternative: shift PVI to 0+148.231	grate 27.822 m (raise 0.082 m)

Topic. Geometry of the symmetrical parabolic vertical curve, and the location of its low point relative to drainage structures.

Key relations. $y = y_{PVC} + g_1 x + \dfrac{A}{200L}x^{2}$; low or high point $x = L|g_1|/A$; PVC $=$ PVI $- L/2$; $K = L/A$; sag headlight control $K = S^{2}/(120+3.5S)$.

Why this works. On a symmetrical parabola the offset from the tangent grows with the square of the distance from the PVC, so the gradient falls linearly from $g_1$ to $g_2$ and reaches zero at the fraction $|g_1|/A$ of the length. That single fact makes the low point a function of only the length and the two grades, which is why lengthening the curve is a legitimate lever: holding the grades and the PVI, the low point retreats towards the PVC at 0.1154 m per metre of added length. The elevation at the low point rises at the same time, which is exactly what allows the drowned inlet to be lifted out of the pond.

Common pitfall. Assuming the low point of a sag curve is at the PVI, or midway along the curve; it is only at mid-length when the two grades are equal and opposite. The second pitfall is to answer with a new curve length and stop — the question asks for the grate elevation, which requires the profile to be re-evaluated on the new parabola rather than on the old one.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §15.5 (vertical curves); TAC, Geometric Design Guide for Canadian Roads, §2.1.3.

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