Question 3 of 7: Flexible pavement design for 10 million ESALs

Question 3: Flexible pavement design for 10 million ESALs (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. A three-layer flexible structure to be proportioned by the AASHTO 1993 structural-number method, with the material and reliability inputs listed below.

Find. The layer coefficients, the required structural number, and a buildable set of layer thicknesses whose provided structural number covers the requirement.

Approach. Read the three layer coefficients and the drainage coefficients from the appendix charts, solve the 1993 flexible equation for the structural number required on the subgrade and again on top of each stiffer layer, then work downward from the surface assigning thicknesses that satisfy both the layered requirement and the minimum-thickness table.

1. Read the layer coefficients. From appendix page 6, the dense-graded asphalt concrete chart at E_AC = 400,000 psi gives $a_1 = 0.42$. The cement-treated base nomograph on the same page, entered at a 7-day unconfined compressive strength of 450 psi, gives $a_2 = 0.17$ and an equivalent base modulus of about $6.4 \times 10^{5}$ psi. From appendix page 7, the granular subbase chart at E_SB = 20,000 psi gives $a_3 = 0.14$, which the published correlation $a_3 = 0.227\log_{10}E_{SB} - 0.839 = 0.137$ confirms.
2. Assign the drainage coefficients. The appendix table is headed Recommended m_i values for untreated base and subbase materials, so it applies to the untreated silty-sand subbase but not to a cement-treated layer. For poor drainage with the structure near saturation 25 % of the time the table gives the lower end of the poor row, $$m_3 = \boxed{0.60}, \qquad m_2 = 1.00\ \text{(cement-treated, not an untreated material)}$$ This is a deliberate engineering judgement and is revisited in the callout below.
3. Solve the design equation for the structural number on the subgrade. The appendix equation is $$\log_{10}W_{18} = Z_R S_0 + 9.36\log_{10}(SN+1) - 0.20 + \frac{\log_{10}\!\left(\dfrac{\Delta PSI}{4.2-1.5}\right)}{0.40 + \dfrac{1094}{(SN+1)^{5.19}}} + 2.32\log_{10}M_R - 8.07$$ Substituting $W_{18} = 10^{7}$, $Z_R S_0 = (-1.282)(0.4) = -0.513$, $\Delta PSI = 1.5$ and $M_R = 7\,000$ psi and solving numerically, $$\boxed{SN_{required} = 5.15}$$
4. Repeat the solution with the stiffer layers as the supporting medium. The layered procedure protects each layer in turn by asking what structural number must sit above it. Entering the same equation with the subbase modulus, $M_R = 20\,000$ psi, gives $SN_2 = 3.49$; entering it with the cement-treated base modulus, $6.4 \times 10^{5}$ psi, gives $SN_1 = 0.77$.
5. Fix the asphalt thickness. The structural need above the base is only $D_1 = SN_1/a_1 = 0.77/0.42 = 1.8$ in., because a cement-treated base is an extremely stiff support. The appendix minimum-thickness table therefore governs: for traffic greater than 7,000,000 ESALs the minimum asphalt concrete is $$D_1 = \boxed{4.0\ \text{in.}} \qquad SN_1^{prov} = 0.42(4.0) = 1.68$$
6. Fix the base thickness. The base must lift the structural number from 1.68 to the 3.49 required on top of the subbase: $$D_2 \ge \frac{SN_2 - SN_1^{prov}}{a_2\,m_2} = \frac{3.49 - 1.68}{0.17(1.00)} = 10.65\ \text{in.} \quad\Longrightarrow\quad D_2 = \boxed{11\ \text{in.}}$$ which comfortably exceeds the 6 in. tabulated minimum. The section now provides $SN = 1.68 + 0.17(11) = 3.55$.
7. Fix the subbase thickness. The subbase must close the remaining gap to 5.15, and it does so at a heavy discount because the drainage coefficient of 0.60 discards 40 % of its contribution: $$D_3 \ge \frac{SN_{req} - SN_{1,2}^{prov}}{a_3\,m_3} = \frac{5.15 - 3.55}{0.14(0.60)} = 19.1\ \text{in.} \quad\Longrightarrow\quad D_3 = \boxed{19.5\ \text{in.}}$$
8. Check the adopted section. Summing the contributions, $$SN_{prov} = 0.42(4.0) + 0.17(11)(1.00) + 0.14(19.5)(0.60) = 1.68 + 1.87 + 1.64 = \boxed{5.19 \ge 5.15}$$ The total bound thickness is 34.5 in., about 876 mm, which is a reasonable structure for 10 million ESALs on a 7,000 psi subgrade with poor drainage.