16-Civ-A6 Highway Design, Construction, and Maintenance · May 2018
Question 6 of 7: Burmister two-layer analysis of plate bearing tests
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2018, 16-Civ-A6 — Highway Design, Construction, and Maintenance. Seven questions of equal value (20 marks each), three hours, closed book, with a ten-page appendix of design charts and tables. Only the first five solutions are marked, but because this set is a study resource all seven questions are solved here.
Unless a question states otherwise the perception–reaction time is taken as $t_{pr}=2.5\ \text{s}$ (the AASHTO design value) under NOTE 2 on page 1, and $g=9.81\ \text{m/s}^{2}$. Stopping and side friction coefficients are read from the “Friction Coefficients to be used in questions” table on appendix page 9; clear-zone widths and their horizontal-curve correction factors come from the two tables on the same page.
Reference texts.
Garber, N.J. and Hoel, L.A., Traffic and Highway Engineering, 5th ed. — Ch. 3 (driver characteristics and stopping sight distance), Ch. 15 (geometric design of highway facilities), Ch. 20 (design of flexible highway pavements).
AASHTO, Guide for Design of Pavement Structures, 1993 — Part II Ch. 2 (flexible pavement design); Figures 2.5–2.7, Table 2.4 and Figure 3.1 are reproduced on appendix pages 2–4.
AASHTO, A Policy on Geometric Design of Highways and Streets (Green Book), 7th ed. — Ch. 3 (sight distance, horizontal and vertical alignment).
Transportation Association of Canada, Geometric Design Guide for Canadian Roads — Ch. 1.2 (design controls), Ch. 2.1 (sight distance), Ch. 3.2 (horizontal alignment and spiral transitions), Ch. 4 (roadside design and clear zones).
Asphalt Institute, Thickness Design — Asphalt Pavements for Highways and Streets (MS-1) — Ch. IV–VI; Design Charts A-1 to A-6 and Tables VI-2, VI-3 are reproduced on appendix pages 5–8.
Huang, Y.H., Pavement Analysis and Design, 2nd ed. — Ch. 2 (stresses and deflections in flexible pavements, Burmister two-layer theory); the $F_{2}$ chart on appendix page 10 is Huang Figure 2.17.
Given. Two plate bearing tests and one design condition, all on the same subgrade:
Test and design data
Case
Loaded area
Load or pressure
Deflection
(a) rigid plate on the subgrade
12 in. diameter ($a=6$ in.)
9,600 lb
0.2 in.
(b) rigid plate on 6 in. of gravel base
12 in. diameter ($a=6$ in.)
16,000 lb
0.2 in.
(c) tyre on the design section
$a=5$ in. circular
100 psi contact pressure
0.1 in. permitted
Poisson ratio (both layers)
$\nu=0.5$
Find. The base-course thickness $h_{1}$ that limits the surface deflection under the design tyre to 0.1 in.
Figure 6.1 — the three cases. (a) fixes the subgrade modulus, (b) fixes the base modulus, (c) is the design condition.
Approach. Case (a) is a rigid plate on a homogeneous half-space, which gives $E_{2}$ in closed form. Case (b) is a rigid plate on a two-layer system, which gives the deflection factor $F_{2}$ and hence $E_{1}/E_{2}$ from the Burmister chart. Case (c) is a flexible loaded area on the same two-layer system: compute the $F_{2}$ the deflection limit allows and read $h_{1}/a$ back off the same curve.
Contact pressure in test (a). The plate is 12 in. in diameter, so $a=6$ in. and$$q_{a}=\dfrac{P}{\pi a^{2}}=\dfrac{9\,600}{\pi(6)^{2}}=84.88\ \text{psi}$$
Subgrade modulus. For a rigid plate on a homogeneous elastic half-space the appendix gives$$w_{0}=\dfrac{\pi(1-\nu^{2})qa}{2E}\qquad\Longrightarrow\qquad E_{2}=\dfrac{\pi(1-\nu^{2})q_{a}a}{2w_{0}}$$With $\nu=0.5$ so that $1-\nu^{2}=0.75$,$$E_{2}=\dfrac{\pi(0.75)(84.88)(6)}{2(0.2)}=\boxed{3\,000\ \text{psi}}$$
Deflection factor in test (b). The same plate now rests on 6 in. of base, so $q_{b}=16\,000/[\pi(6)^{2}]=141.47$ psi and the rigid-plate two-layer expression applies:$$w_{0}=\dfrac{1.18\,q a}{E_{2}}F_{2}\qquad\Longrightarrow\qquad F_{2}=\dfrac{w_{0}E_{2}}{1.18\,q_{b}a}=\dfrac{0.2(3\,000)}{1.18(141.47)(6)}=\boxed{0.599}$$Note that the coefficient 1.18 is just $\pi(1-\nu^{2})/2=1.178$ evaluated at $\nu=0.5$, so the two formulae are the same statement.
Modular ratio from the Burmister chart. The thickness ratio in test (b) is $h_{1}/a=6/6=1.0$. Entering the $F_{2}$ chart on appendix page 10 at $(h_{1}/a,\ F_{2})=(1.0,\ 0.599)$ lands on the second curve from the top, so$$\dfrac{E_{1}}{E_{2}}=5\qquad\Longrightarrow\qquad E_{1}=5(3\,000)=\boxed{15\,000\ \text{psi}}$$This is a sensible gravel-base stiffness, which is a useful sanity check on the chart read.
Deflection factor the design case allows. A tyre prints a flexible, not a rigid, loaded area, so the 1.5 coefficient applies:$$w_{0}=\dfrac{1.5\,q a}{E_{2}}F_{2}\qquad\Longrightarrow\qquad F_{2}=\dfrac{w_{0}E_{2}}{1.5\,qa}=\dfrac{0.1(3\,000)}{1.5(100)(5)}=\dfrac{300}{750}=\boxed{0.400}$$
Required thickness. Following the same $E_{1}/E_{2}=5$ curve to $F_{2}=0.400$ gives $h_{1}/a\approx 2.3$, and the design loaded radius is $a=5$ in., so$$h_{1}=2.3\times 5=\boxed{11.5\ \text{in.}\ (292\ \text{mm})}$$Specify 300 mm (12 in.) of the same gravel base course. The thickness roughly doubles relative to the 6 in. test layer chiefly because the permitted deflection has been halved (0.1 in. against 0.2 in.), and because a flexible tyre load deflects the surface about 27 % more than a rigid plate at the same pressure.
Check: the source disagrees with itself on the design contact pressure. The text of Question 6 specifies “a tire exerting a contact pressure of 100 psi”, while the label on Figure 3-c reads “80 psi”. Both readings are as printed, so this is a genuine inconsistency in the paper. The question text governs and the solution above uses 100 psi. For completeness, at 80 psi the required factor becomes $F_{2}=0.1(3\,000)/[1.5(80)(5)]=0.500$, which the same curve meets at $h_{1}/a\approx 1.45$, i.e. $h_{1}=7.3$ in. (185 mm). An examination answer should state the discrepancy, choose one reading and solve it consistently, exactly as NOTE 1 on page 1 invites.
