16-Civ-B7 Transportation Planning and Engineering · May 2013
Question 1 of 7: Earthwork Volumes — Reservoir Storage and Embankment Fill
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Civ-B7 Highway Engineering, National Examinations May 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 on the last page
confirms 20 marks per question: Q1 (a) and (b) 10 marks each; Q2 (a) through (e)
4 marks each; Q3 (a) to (j) 2 marks each; Q4 (a) and (b) 10 marks each; Q5 (a) and (b)
10 marks each; Q6 (a) through (e) 4 marks each; Q7 20 marks. All seven
printed questions are worked here, 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 not given but required may be assumed, and that
assumptions should be recorded with the answer — several questions below need
that licence, and each assumption is flagged where it is made.
Reference texts. N.J. Garber and L.A. Hoel, Traffic and Highway
Engineering, 5th ed. (geometric design, sight distance, vertical curves, earthwork,
pavement design); AASHTO, Guide for Design of Pavement Structures (1993)
(rigid and flexible thickness design, reliability, drainage and load-transfer
coefficients); Transportation Association of Canada, Geometric Design Guide for
Canadian Roads (Canadian design-domain values for sight distance and vertical
curvature); Asphalt Institute, Mix Design Methods MS-2, 7th ed. (mixture
volumetrics, VMA, VFA, absorbed binder); B.M. Das, Principles of Geotechnical
Engineering, 9th ed. (compaction, Proctor testing, zero-air-voids line, CBR);
M.S. Mamlouk and J.P. Zaniewski, Materials for Civil and Construction Engineers,
4th ed. (concrete and asphalt materials); A.M. Neville, Properties of Concrete,
5th ed., and CSA A23.1 (air entrainment, curing, joints in concrete pavement).
Check — assumptions carried through this paper. Three items
are not supplied by the exam and are assumed under the paper’s own Note 2
(“any data, not given but required, can be assumed”), each stated again at
the point of use: (i) Question 5 gives the mass of the Proctor mould but not its
volume, so the ASTM D698 / AASHTO T99 standard 101.6 mm mould volume of
944 cm3 is used; (ii) Question 6 does not name a design speed, so the
available stopping sight distance is computed from the Canadian/AASHTO eye and object
heights of 1.08 m and 0.60 m; (iii) Question 7 lists the modulus of subgrade reaction
as “1.0 MPa”, which is dimensionally incomplete — it is read as
1.0 MPa/m and the sensitivity of the answer to that reading is reported with the
result.
Question 1: Earthwork Volumes — Reservoir Storage and Embankment Fill (20 marks)
Given. A pond with straight sloping sides rising 20 m from a
100 m by 50 m rectangular floor to a 200 m by 100 m rectangular rim, plus three
planimetered embankment end areas spaced 1000 m apart along the centre line.
Given data
Item
Value
Pond bottom elevation and plan size
100.000 m; 100 m × 50 m
Pond top elevation and plan size
120.000 m; 200 m × 100 m
Side geometry
straight (plane) sloping sides
Embankment end areas
640, 1500 and 800 m2
Station interval
1000 m
Find. The plan dimensions at elevations 105.000, 110.000 and
115.000 m and the resulting storage capacity of the pond; then the fill volume between
stations 50+000 and 52+000 by both the average-end-area and prismoidal rules.
Figure 1.1 — Pond geometry: plane side slopes make every plan dimension a linear function of elevation, so the contour areas at 105, 110 and 115 m follow by interpolation.
Approach. Because the sides are plane surfaces, both plan
dimensions vary linearly with elevation, so the contour dimensions follow by
straight-line interpolation; the contour areas are then combined vertically by the
average-end-area (trapezoidal) rule and by the prismoidal (Simpson) rule, and the same
two rules are applied horizontally to the embankment end areas.
(a) Interpolate the plan dimensions. Let $f$ be the fraction of the
total height reached at elevation $z$:
$$f=\frac{z-100}{120-100},\qquad L=100+100f\ \text{m},\qquad W=50+50f\ \text{m}$$
Both the length and the width double over the 20 m rise, so every horizontal section
stays geometrically similar to the floor — a useful check on the arithmetic.
Contour dimensions and areas. Evaluating at the three requested
elevations, and adding the floor and rim for completeness:
$$\begin{aligned}
z=105\ \text{m}\ (f=0.25):&\quad 125.0\ \text{m}\times 62.5\ \text{m},\ A=7\,812.5\ \text{m}^2\\
z=110\ \text{m}\ (f=0.50):&\quad 150.0\ \text{m}\times 75.0\ \text{m},\ A=11\,250.0\ \text{m}^2\\
z=115\ \text{m}\ (f=0.75):&\quad 175.0\ \text{m}\times 87.5\ \text{m},\ A=15\,312.5\ \text{m}^2
\end{aligned}$$
with $A_{100}=100\times 50=5\,000\ \text{m}^2$ at the floor and
$A_{120}=200\times 100=20\,000\ \text{m}^2$ at the rim.
Storage by the average end area (trapezoidal) rule. With five
contour areas at a uniform vertical interval $h=5$ m,
$$V=h\left[\frac{A_0+A_4}{2}+A_1+A_2+A_3\right]
=5\left[\frac{5\,000+20\,000}{2}+7\,812.5+11\,250+15\,312.5\right]$$
$$V=5\,(12\,500+34\,375)=\boxed{234\,375\ \text{m}^3}$$
Storage by the prismoidal (Simpson) rule. Four intervals is an
even number, so Simpson’s rule applies directly:
$$V=\frac{h}{3}\left[A_0+4A_1+2A_2+4A_3+A_4\right]
=\frac{5}{3}\left[5\,000+31\,250+22\,500+61\,250+20\,000\right]$$
$$V=\frac{5}{3}(140\,000)=\boxed{233\,333\ \text{m}^3}$$
Which figure is the capacity? The pond is a frustum of a pyramid,
whose area varies as a quadratic function of elevation, and Simpson’s rule is
exact for a quadratic. Integrating directly confirms it:
$$V=\int_{100}^{120}A(z)\,dz=20\int_0^1 5\,000(1+f)^2\,df
=100\,000\cdot\frac{2^3-1^3}{3}=233\,333.3\ \text{m}^3$$
so the prismoidal value is the true capacity and the average-end-area figure
overstates it by 1 042 m3, about 0.45 %. Quote
233 333 m3 (≈ 2.33 × 105 m3)
as the storage capacity.
Figure 1.2 — The three planimetered end areas along the embankment. The middle section is a pronounced peak, which is what separates the two volume rules.
(b)(i) Embankment fill — average end area method. Each 1000 m bay is
treated as a prism whose area is the mean of its two end areas:
$$V=\frac{L}{2}\big[(A_1+A_2)+(A_2+A_3)\big]
=\frac{1000}{2}\big[(640+1500)+(1500+800)\big]$$
$$V=500\,(2\,140+2\,300)=\boxed{2\,220\,000\ \text{m}^3}$$
(b)(ii) Embankment fill — prismoidal formula. The two bays together
form a single prismoid of total length $L_t=2000$ m with 1500 m2 as the
middle area:
$$V=\frac{L_t}{6}\left[A_1+4A_m+A_2\right]=\frac{2000}{6}\left[640+4(1500)+800\right]$$
$$V=333.33\,(7\,440)=\boxed{2\,480\,000\ \text{m}^3}$$
Interpret the difference. Here the prismoidal volume is the larger
of the two, by 260 000 m3 or 11.7 %, because the middle section is a
pronounced peak rather than a value part-way between its neighbours; the
average-end-area rule, which ignores the middle section entirely when the bays are
taken pairwise, cannot see that bulge. The prismoidal figure is the one to carry into a
mass-haul diagram or a payment quantity.
The two parts illustrate the same rule pulling in opposite directions: on the pond,
where the area function is convex upward, average-end-area is generous, and on the
embankment, where the area function peaks in the middle, it is conservative. That is
not a contradiction but the signature of a second-order correction, and it is the
reason the prismoidal rule is the one specified for contract quantities on Canadian
highway work whenever the sections are widely spaced or an area changes sharply between
them, as it does here; at the usual 20 to 30 m section spacing the two rules differ by
so little that average end area is accepted.
