Question 5 of 7: Plain jointed concrete slab thickness for 12 million ESALs

Question 5: Plain jointed concrete slab thickness for 12 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 plain jointed concrete pavement on a 10 in. unbound granular subbase over a high-plasticity clay subgrade with bedrock only six feet down.

Find. The design slab thickness D.

Approach. Build the effective modulus of subgrade reaction in three steps — composite k for the subbase, correction for the shallow rigid foundation, correction for loss of support — select the load transfer and drainage coefficients from the appendix tables, then solve the 1993 rigid design equation for D.

1. Find the composite modulus of subgrade reaction. Appendix page 9 converts a subgrade modulus, a subbase modulus and a subbase thickness into a composite k on the assumption of semi-infinite subgrade depth. Entering M_R = 5,000 psi, E_SB = 30,000 psi and D_SB = 10 in. gives $$k_{\infty} \approx \boxed{390\ \text{pci}}$$ The chart is a two-layer plate-deflection solution, and its own no-subbase rule $k = M_R/19.4 = 258$ pci reproduces the left-hand edge of the same chart, which is a useful check that the entry point is right: the 10 in. of granular subbase is worth about half again on the foundation stiffness.
2. Correct for the rigid foundation near the surface. The chart above assumes rock is more than 10 ft below the subgrade surface. Here bedrock is at 6 ft, so the deflection basin is truncated and the foundation behaves stiffer. Entering the second chart on appendix page 9 with M_R = 5,000 psi, a depth to rigid foundation of 6 ft and k_∞ = 390 pci, $$k = \boxed{570\ \text{pci}}$$
3. Correct for potential loss of support. An unbound granular subbase over a high-plasticity clay is exactly the case AASHTO assigns a loss-of-support factor LS = 1.0, because pumping and erosion at joints will eventually leave voids under the slab corners. Appendix page 10 converts the elastic k to an effective k on that basis; the chart’s own worked example (540 pci reduced to 170 pci) fixes the curve, and applying it here, $$k_{eff} = \boxed{178\ \text{pci}}$$ This is by far the largest of the three corrections and it is the one candidates most often omit: it more than undoes the benefit of both the subbase and the shallow bedrock.
4. Select the load transfer coefficient. From the appendix table for a plain jointed pavement with load transfer devices present and an asphalt shoulder, $$J = \boxed{3.2}$$ Had tied concrete shoulders been used instead, J would fall to 2.5 to 3.1 and the slab would thin appreciably; the asphalt shoulder is the more onerous case because the slab edge is unsupported.
5. Select the drainage coefficient. From the rigid-pavement drainage table, good drainage with the structure near saturation 15 % of the time gives the range 1.10 to 1.00; taking the midpoint, $$C_d = \boxed{1.05}$$
6. Write the rigid design equation. The 1993 rigid equation is $$\log_{10}W_{18} = Z_R S_0 + 7.35\log_{10}(D+1) - 0.06 + \frac{\log_{10}\!\left(\dfrac{\Delta PSI}{4.5-1.5}\right)}{1 + \dfrac{1.624\times10^{7}}{(D+1)^{8.46}}} + (4.22 - 0.32\,p_t)\log_{10}\!\left[\frac{S'_c\,C_d\,(D^{0.75} - 1.132)}{215.63\,J\left(D^{0.75} - \dfrac{18.42}{(E_c/k)^{0.25}}\right)}\right]$$ Note that the rigid form is built around an initial serviceability of 4.5, which is exactly what the question supplies, so no adjustment of the denominator is needed.
7. Solve for the slab thickness. Substituting $W_{18} = 1.2\times10^{7}$, $Z_R S_0 = -0.385$, $\Delta PSI = 2.0$, $p_t = 2.5$, $S'_c = 600$ psi, $C_d = 1.05$, $J = 3.2$, $E_c = 3.6\times10^{6}$ psi and $k = 178$ pci, and iterating on D, $$\boxed{D = 10.1\ \text{in.} \quad\Longrightarrow\quad \text{specify } D = 10.5\ \text{in.}\ (267\ \text{mm})}$$ Substituting 10.5 in. back into the equation returns an allowable traffic of 1.5 × 10⁷ ESALs against the 1.2 × 10⁷ required, so the specified slab carries the design traffic with a margin.
8. Test how much the foundation assumption matters. Because the radius of relative stiffness varies as k to the power of one quarter, the slab thickness is remarkably insensitive to the foundation. Repeating the solution at k = 120 pci gives D = 10.31 in. and at k = 300 pci gives D = 9.83 in. — a 150 % change in k moves the answer by less than half an inch. The design conclusion, a 10.5 in. slab, is therefore robust against the reading taken from the two printed k charts, and the marks are properly in the chain of corrections rather than in the third decimal of any one of them.