Question 1 of 5: AASHTO-93 design of a plain jointed concrete pavement

Question 1: AASHTO-93 design of a plain jointed concrete pavement (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 jointed plain concrete pavement (JPCP) on 12 in of unbound granular subbase over a high-plasticity clay subgrade, with bedrock only five feet below the subgrade surface.

Find. The required thickness D of the concrete slab.

Approach. Build the effective modulus of subgrade reaction in three chart steps (composite k for the subbase, correction for the shallow bedrock, correction for loss of support), select the drainage and load-transfer coefficients from Tables 2.5 and 2.6, and then solve the AASHTO-93 rigid performance equation for the slab thickness.

1. Fix the design inputs that come straight from the tables. The serviceability loss is the difference of the two printed indices, $$\Delta PSI = p_i - p_t = 4.5 - 2.5 = 2.0$$ At 90 % reliability the standard normal deviate is $Z_R = -1.282$, and the exam gives $S_0 = 0.30$ (inside the 0.30–0.40 band the appendix lists for rigid pavements). Table 2.5 is entered with good drainage and 5–25 % of the time saturated, which brackets the drainage coefficient between 1.10 and 1.00; the mid-range value $C_d = 1.05$ is adopted. Table 2.6 is entered with a plain jointed pavement, load-transfer devices present (the dowels) and an asphalt shoulder, which gives the load transfer coefficient $J = 3.2$ directly.
2. Convert the subgrade and subbase into one composite modulus of subgrade reaction. With no subbase at all the appendix rule would give $$k_{\infty} = \frac{M_R}{19.4} = \frac{4000}{19.4} = 206\ \text{pci}$$ Figure 3.3 adds the 12 in of 30,000 psi granular subbase. Entering the chart with $E_{SB} = 30{,}000\ \text{psi}$, $D_{SB} = 12\ \text{in}$ and $M_R = 4{,}000\ \text{psi}$, and reading across the turning line, $$\boxed{k_{\infty} \approx 363\ \text{pci}}$$ so the subbase is worth about a 76 % increase in foundation stiffness.
3. Correct for the rigid foundation five feet down. Figure 3.4 applies whenever bedrock lies within 10 ft of the subgrade surface, which is exactly this case. Entering at $M_R = 4{,}000\ \text{psi}$, rising to the $D_{SG} = 5\ \text{ft}$ curve, transferring across to the $k_{\infty} = 363\ \text{pci}$ contour and dropping to the axis gives $$k \approx 800\ \text{pci}$$ The correction is large because a soft clay only five feet thick cannot deflect the way a semi-infinite clay can. Ignoring Figure 3.4 here would leave the design about 0.3 in thicker than necessary — conservative, but it would not be the AASHTO answer.
4. Take the loss of subbase support back off again. The subbase is unbound granular (E between 15,000 and 45,000 psi), for which the appendix table gives a loss-of-support factor $LS = 1.0$. Figure 3.6 on the $LS = 1.0$ line — anchored on the chart’s own printed example, which maps 540 pci onto 170 pci — reduces 800 pci to $$\boxed{k_{\text{eff}} \approx 237\ \text{pci}}$$ This step matters more than either of the two before it: pumping and erosion of an unbound granular layer under a jointed slab throw away most of what the subbase and the shallow bedrock contributed.
5. Solve the AASHTO-93 rigid performance equation for the slab thickness. The equation printed on appendix pages 5 and 8 is $$\log_{10}W_{18} = Z_R S_0 + 7.35\log_{10}(D+1) - 0.06 + \frac{\log_{10}\!\left[\dfrac{\Delta PSI}{4.5-1.5}\right]}{1 + \dfrac{1.624\times10^{7}}{(D+1)^{8.46}}} + (4.22 - 0.32\,p_t)\log_{10}\!\left[\frac{S'_c\,C_d\left(D^{0.75}-1.132\right)}{215.63\,J\left(D^{0.75} - \dfrac{18.42}{(E_c/k)^{0.25}}\right)}\right]$$ Substituting $W_{18} = 10^{7}$, $Z_R S_0 = -1.282 \times 0.30$, $\Delta PSI = 2.0$, $p_t = 2.5$, $S'_c = 650$, $C_d = 1.05$, $J = 3.2$, $E_c = 3.8\times10^{6}$ and $k = 237\ \text{pci}$, and solving for D by iteration, $$\boxed{D_{\text{required}} = 9.28\ \text{in}}$$
6. Round up to a constructible thickness and confirm the margin. Slabs are specified in half-inch increments, so take $D = 9.5\ \text{in}\ (240\ \text{mm})$. Substituting 9.5 in back into the same equation returns an allowable $$W_{18} = 1.15\times10^{7}\ \text{ESAL}$$ against the 1.00 × 10⁷ demanded — a 15 % surplus, which is the right side of the line to be on.
7. Check how much the answer depends on the graphical steps. Slab thickness varies with k as roughly $k^{-1/4}$ through the radius of relative stiffness, so the chart reads are far less critical than they look. Re-solving with the Figure 3.4 value swept over a factor of 2.5 gives:

   Sensitivity of the required slab thickness to the Figure 3.4 read

   k from Fig 3.4 (pci)	k after LS = 1.0 (pci)	Required D (in)
   400	132	9.58
   600	186	9.42
   800 (adopted)	237	9.28
   1000	286	9.17

   The whole range collapses onto the same specified 9.5 in slab, so the recommendation is robust against any reasonable reading of the nomograph.