16-Civ-A6 Highway Design, Construction, and Maintenance · December 2018

Question 3 of 5: Rigid pavement — back-calculating the design ESAL and revising the slab

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 3: Rigid pavement — back-calculating the design ESAL and revising the slab (25 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 plain jointed concrete pavement whose preliminary design returned a 9 in. slab, together with every input used in that design, and a subsequent borehole finding that a rigid layer lies about 2 ft below the subgrade surface.

Find. (a) the cumulative 18-kip applications $W_{18}$ that the 9 in. slab was designed to carry; (b) the slab thickness that the same traffic requires once the shallow rigid layer is taken into account.

Approach. Build the support term first — composite modulus of subgrade reaction from Figure 3.3, then the loss-of-support correction from Figure 3.6 — and read the load transfer and drainage coefficients from Tables 2.6 and 2.5. With every term of the rigid design equation then known except $W_{18}$, part (a) is a direct evaluation at $D = 9$ in. Part (b) repeats the support calculation with the rigid layer present and solves the same equation for $D$ at the traffic found in part (a).

1. Read the composite modulus of subgrade reaction. Figure 3.3 combines the subbase and the roadbed into a single spring constant. Entering the subbase-thickness scale at $D_{SB} = 12$ in., rising to the $E_{SB} = 30{,}000$ psi curve, and dropping to the $M_R = 5{,}000$ psi curve and across the turning line gives $$k_{\infty} \approx 470\ \text{pci}$$ As a check on the reading, the appendix also states $k = M_R/19.4 = 5000/19.4 = 258$ pci when no subbase is used, so the 12 in. granular layer is credited with raising the support by a factor of about 1.8 — a physically sensible increase for a stiff layer that thick.
2. Correct for potential loss of subbase support. The subbase is an unbound granular material with $E_{SB} = 30{,}000$ psi, which the appendix table places in the "Unbound Granular Materials (15,000 to 45,000 psi)" row with $LS = 1.0$ to $3.0$. A properly graded, well-compacted dense granular subbase under a dowelled jointed pavement warrants the favourable end of that range, so $LS = 1.0$ is adopted. Entering Figure 3.6 on its $LS = 1.0$ line — the line through the chart's own printed example, which maps 540 pci onto 170 pci — at $k_{\infty} = 470$ pci gives $$\boxed{k_{\text{effective}} \approx 150\ \text{pci}}$$ The correction is severe, and deliberately so: it represents the erosion and pumping of fines from beneath the slab edges over the life of the pavement, and it is the value with the erosion allowance, not the as-built value, that belongs in the design equation.
3. Read the load transfer coefficient. Table 2.6 is entered for a plain jointed pavement, with load transfer devices present (the question states dowel bars at the transverse joints and tie bars at the longitudinal joints) and asphalt shoulders, which gives a single tabulated value: $$J = 3.2$$
4. Read the drainage coefficient. Table 2.5 is entered on the "Good" row in the 5–25 % column, which spans 1.10 to 1.00. The stated 15 % sits close to the middle of that band, so $$C_d = 1.05$$
5. Complete the remaining inputs. At $R = 95\ \%$ the standard normal deviate is $Z_R = -1.645$; $S_o = 0.35$ as stated. The rigid design equation is written about an initial serviceability of 4.5 — the value used at the AASHO Road Test and the value embedded in the $4.5-1.5$ denominator printed on appendix page 8 — so $$\Delta PSI = p_0 - p_t = 4.5 - 2.5 = 2.0$$
6. Evaluate the rigid design equation at the preliminary thickness. The equation reproduced on appendix page 8 is $$\log_{10}W_{18} = Z_R S_o + 7.35\log_{10}(D+1) - 0.06 + \frac{\log_{10}\!\left(\dfrac{\Delta PSI}{4.5-1.5}\right)}{1 + \dfrac{1.624\times10^{7}}{(D+1)^{8.46}}} + (4.22 - 0.32p_t)\log_{10}\!\left[\frac{S'_c\,C_d\,(D^{0.75} - 1.132)}{215.63\,J\left(D^{0.75} - \dfrac{18.42}{(E_c/k)^{0.25}}\right)}\right]$$ Every term on the right is now known at $D = 9$ in., so the equation is evaluated directly rather than solved. Substituting $Z_R S_o = -0.5758$, $D+1 = 10$, $S'_c = 700$, $C_d = 1.05$, $J = 3.2$, $E_c/k = 4\times10^{6}/150 = 26\,667$ and $p_t = 2.5$ returns $\log_{10}W_{18} = 6.759$, that is $$\boxed{W_{18} = 5.74 \times 10^{6}\ \text{18-kip ESAL applications}}$$ Solving the same equation the other way — for $D$ at $W_{18} = 5.74\times10^{6}$ — returns 9.00 in., which confirms the evaluation is self-consistent.
7. Re-establish the support with the rigid layer present. The caption of Figure 3.3 is explicit that it "assumes a semi-infinite subgrade depth", and defines semi-infinite as more than ten feet below the subgrade surface. A rigid layer two feet down violates that assumption badly: the compressible soil that would otherwise deflect under the slab is only 610 mm thick, so the same load produces a much smaller deflection and the support is stiffer. AASHTO handles this with Figure 3.4, the rigid-foundation correction, which is not reproduced in this paper's ten-page appendix. Invoking Note 2, the correction is assumed: at $M_R = 5{,}000$ psi and a depth to rigid foundation of 2 ft, the chart approximately doubles the composite value, so $k_{\infty} \approx 900$ pci is taken forward. Applying the same $LS = 1.0$ correction from Figure 3.6 to that value gives $$k_{\text{effective, revised}} \approx 260\ \text{pci}$$
8. Solve for the revised slab thickness. Holding $W_{18} = 5.74\times10^{6}$ and every other input unchanged, and solving the same equation for $D$ with $k = 260$ pci: $$\boxed{D = 8.70\ \text{in.} \;\Rightarrow\; \text{specify an 8.5 in. (215 mm) slab}}$$ rounding to the nearest half-inch as the appendix example does.
9. Test how much the answer depends on the assumed correction. Because Figure 3.4 had to be assumed, the honest way to present part (b) is to show the whole family of answers:

   Slab thickness required at $W_{18} = 5.74\times10^{6}$ as the support varies

   Effective $k$ (pci)	150	200	260	350	500
   Required $D$ (in.)	9.00	8.85	8.70	8.51	8.22

   Tripling the modulus of subgrade reaction, from 150 to 500 pci, thins the slab by less than 0.8 in. Whatever value Figure 3.4 would have returned, the conclusion is the same: the shallow rigid layer allows the slab to come down by roughly half an inch, from 9 in. to 8.5 in., and no more.
10. Report the engineering consequence, not only the number. A half-inch of concrete over a project of any size is worth taking, but the far more important consequence of the borehole finding is not structural. A rigid layer two feet down means the subgrade cannot drain vertically, so water entering the granular subbase will perch on the rock; that attacks the very loss-of-support assumption ($LS = 1.0$) on which Step 2 rests, and it makes frost-heave differential across the site more likely where the rock surface undulates. The recommendation that should accompany the revised thickness is therefore a longitudinal subdrain at the subbase level and confirmation that the 2 ft is an average rather than a minimum — if the rock rises to within a few inches of the subbase anywhere along the alignment, that location will control performance regardless of slab thickness.