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

Question 7 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 7 20 marks — (a) 8, (b) 12

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. Percent passing for two aggregate stockpiles and the specification band they must be blended to satisfy:

Given data — percent passing
Sieve size	19 mm	12.5 mm	9.5 mm	4.75 mm	2.36 mm	0.60 mm	0.30 mm	0.15 mm	0.075 mm
Specification limits	100	80–100	75–95	55–65	35–45	25–35	15–25	10–15	5–10
Aggregate A	100	100	95	70	50	40	30	20	10
Aggregate B	100	90	60	30	10	0	0	0	0

Find. The minimum and maximum proportions of aggregates A and B for which every sieve of the blend falls inside the specification band, a selected working blend, and a semi-log plot of A, B, the blend and the specification limits.

Approach. Write the blend as a linear combination of the two stockpiles, invert that relation on each sieve to obtain the interval of proportions that keeps that sieve inside its band, and intersect the nine intervals; the governing constraints are then whichever sieves produce the highest lower bound and the lowest upper bound.

Part (a) — workable range of proportions

Write the blend equation. If a fraction $p$ of aggregate A is combined with $(1-p)$ of aggregate B, then on any sieve the percent passing of the blend is the weighted mean of the two stockpiles: $$P = p\,A + (1-p)\,B = B + p\,(A - B)$$ The relation is linear in $p$ on every sieve, which is what makes both the algebraic and the graphical (Rothfuchs) solutions possible.
Invert the blend equation on each sieve. Requiring $L \le P \le U$ for the lower and upper specification limits, and noting that $A > B$ on every sieve where the two differ, $$\frac{L - B}{A - B} \;\le\; p \;\le\; \frac{U - B}{A - B}$$ Applied to the 4.75 mm sieve, where $A = 70$, $B = 30$ and the band is 55 to 65, $$p \ge \frac{55 - 30}{70 - 30} = 0.625, \qquad p \le \frac{65 - 30}{70 - 30} = 0.875$$

Repeat on every sieve and collect the intervals. The 19 mm sieve imposes no constraint, since both stockpiles pass 100% and the specification requires 100%. The remaining eight give:

Permissible fraction of aggregate A, sieve by sieve
Sieve (mm)	$A$</th><th>$B$</th><th>Band</th><th>Lower bound on $p$</th><th>Upper bound on $p$
19	100	100	100	no constraint ($A = B$)
12.5	100	90	80–100	0.000 (any)	1.000
9.5	95	60	75–95	0.429	1.000
4.75	70	30	55–65	0.625	0.875
2.36	50	10	35–45	0.625	0.875
0.60	40	0	25–35	0.625	0.875
0.30	30	0	15–25	0.500	0.833
0.15	20	0	10–15	0.500	0.750
0.075	10	0	5–10	0.500	1.000

Intersect the intervals to obtain the workable range. The binding lower bound is the largest of the lower bounds and the binding upper bound is the smallest of the upper bounds: $$p_{min} = \max(0.429, 0.625, 0.625, 0.625, 0.500, 0.500, 0.500) = 0.625$$ $$p_{max} = \min(1.000, 1.000, 0.875, 0.875, 0.875, 0.833, 0.750, 1.000) = 0.750$$ so that $$\boxed{62.5\% \le A \le 75.0\%}\qquad\text{and correspondingly}\qquad \boxed{25.0\% \le B \le 37.5\%}$$
Identify which sieves govern, and why. Three sieves — 4.75 mm, 2.36 mm and 0.60 mm — produce the same lower bound of 0.625, because on each of them the difference $A - B$ is exactly 40 percentage points and the lower limit sits 25 points above B. The upper bound comes from the 0.150 mm sieve alone: aggregate B contains nothing finer than 0.600 mm, so every particle passing 0.150 mm in the blend comes from A, and the 15% ceiling on that sieve caps A at $15/20 = 75\%$. The workable window is therefore only 12.5 percentage points wide, which is narrow; the two stockpiles are only marginally compatible with this specification.
Select a working blend and prove it. A designer takes a round proportion near the middle of the window — the midpoint is 68.75% — so select $$\boxed{70\%\ \text{aggregate A} + 30\%\ \text{aggregate B}}$$ Evaluating $P = 0.70A + 0.30B$ on each sieve, for instance at 2.36 mm, $$P_{2.36} = 0.70(50) + 0.30(10) = 35 + 3 = 38\%$$ against a band of 35 to 45. The complete check is tabulated in the results below; every sieve falls inside its limits, with the tightest margins at 2.36 mm (3 points above the floor) and 0.15 mm (1 point below the ceiling).

Part (b) — semi-log gradation plot

Plot all four curves on the semi-log chart. Percent passing is the linear ordinate and sieve size the logarithmic abscissa. The specification limits are plotted as an upper and a lower curve enclosing a band; aggregate A, aggregate B and the 70/30 blend are plotted as separate curves through the tabulated points.
Read the plot. Aggregate A lies above the band throughout the middle of the range — it is too fine on its own, exceeding the upper limit at 4.75 mm (70 against 65) and at 2.36 mm (50 against 45). Aggregate B lies below the band from 9.5 mm down — too coarse on its own, with no material at all finer than 0.600 mm. Neither stockpile is usable alone, but the blend curve threads the band on every sieve, which is the visual statement of the algebra in part (a).
Note what the plot adds beyond the numbers. The blend curve runs close to the upper limit at the coarse end and close to the lower limit at the fine end, so it is a slightly coarse-of-centre mixture. If the plant experienced normal stockpile variation of a few percent on the 0.150 mm sieve, the blend could drift above the 15% ceiling; the graph shows at a glance where the quality-control attention should go, and it argues for keeping the A fraction nearer 68% than 72% in production.

Semi-log gradation chart: aggregates A and B, the selected 70/30 blend and the specification band.

Final results — Question 7
Sieve (mm)	Aggregate A (%)	Aggregate B (%)	Blend 70A/30B (%)	Specification band (%)	Within band?
19	100	100	100.0	100	yes
12.5	100	90	97.0	80–100	yes
9.5	95	60	84.5	75–95	yes
4.75	70	30	58.0	55–65	yes
2.36	50	10	38.0	35–45	yes
0.60	40	0	28.0	25–35	yes
0.30	30	0	21.0	15–25	yes
0.15	20	0	14.0	10–15	yes
0.075	10	0	7.0	5–10	yes
Permissible range: aggregate A from 62.5% to 75.0%; aggregate B from 25.0% to 37.5%. Governing sieves: 4.75, 2.36 and 0.60 mm set the minimum A; 0.15 mm sets the maximum A.

Topic. Blending two aggregate stockpiles to a specification band, algebraically and on a semi-log gradation chart.

Key relations. The blend equation $P = pA + (1-p)B$; its inversion to a permissible interval on each sieve, $(L-B)/(A-B) \le p \le (U-B)/(A-B)$ when $A > B$, with the inequalities reversed when $A < B$; the workable range as the intersection $p_{min} = \max\{$lower bounds$\}$, $p_{max} = \min\{$upper bounds$\}$; and, for three stockpiles, the same idea extended to a system in two independent proportions.

Why this works. Percent passing is an additive property of a mixture on a mass basis: if 70% of the mass comes from a stockpile of which half passes 2.36 mm, that stockpile contributes 35 points of the blend’s passing on that sieve. Because the contribution is linear in the proportion, each specification band translates into a straight interval in $p$, and the feasible set is the intersection of nine intervals on the line — which is either an interval or empty. If it were empty, no two-stockpile blend could meet the specification and a third material would be required. The graphical Rothfuchs construction is the same computation performed by straightedge instead of arithmetic.

Common pitfall. The commonest mistake is stopping at the first sieve that gives a constraint instead of intersecting all of them, which typically produces a range far wider than the true one. The second is losing track of which inequality reverses when $B > A$ on a sieve; the safest habit is to compute both endpoints and then sort them. The third is reporting only the range and never selecting and verifying a specific blend — the question asks for proportions, but a complete answer chooses one, tabulates the resulting gradation, and shows that every sieve passes.

Source. Asphalt Institute, Mix Design Methods for Asphalt Concrete MS-2, aggregate blending and gradation control; M.S. Mamlouk and J.P. Zaniewski, Materials for Civil and Construction Engineers, 4th ed., Chapter 5, blending aggregates to meet a specification; N.J. Garber and L.A. Hoel, Traffic and Highway Engineering, 5th ed., Chapter 19.

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