# Step-by Solution Let's break the problem and the AASO flexible pavement design equation to solve for the probability the pavement retains a PSI above 2.5 after 25 years. --- ## **Given Data** - **Wearing Surface:** 4 in. sand-mix asphalt - **Base:** 10 in. soil - **Subbase:** 10 in. crushed stone - **Traffic:** - 400 single axles (20 kip) per day - 900 tandem axles (35 kip) per day - **Subgrade CBR:** 8 - **Overall Standard Deviation, $S_o$:** .45 - **Initial Serviceability Index, $P_$:** 4.2 - **Terminal Serviceability Index, $P_t$:** 2.5 - **Drainage Coefficient:** 1. - **Design Life:** 25 years --- ## **1. Calculate Structural Number (SN)** ### **Layer Coefficients** Assume typical values (AASHTO Guide): - Asphalt (a₁): .44 - Soil cement (a₂): .14 - Crushed stone (a₃): .11 **Drainage Coefficient ($m$):** 1. for all layers ### **SN Formula** \[ SN = a_1 D_1 + a_2 D_2 m_2 + a_3 D_3 m_3 \] Where: - $D_1$, $D_2$, $D_3$ are layer thicknesses in inches \[ SN = .44 \times 4 + .14 \times 10 \times 1. + .11 \times 10 \times 1. \] \[ SN = 1.76 + 1.4 + 1.1 = 4.26 \] --- ## **2. Convert Axle Loads to 18-kip ESALs** ### **AASHTO Load Equivalency Factors (Approximate):** - **20-kip single axle:** LEF ≈ 1.2 - **35-kip tandem axle:** LEF ≈ 1.8 ### **Daily ESALs:** - **Single:** $400 \times 1.2 = 480$ - **Tandem:** $900 \times 1.8 = 162$ - **Total per day:** $480 + 162 = 210$ ESALs/day ### **25-year ESALs:** \[ ESAL_{total} = 210 \times 365 \times 25 = 19,162,500 \approx 19.2 \text{ million} \] --- ## **3. AASHTO Pavement Design Equation** \[ \log_{10}(W_{18}) = Z_R S_o + 9.36 \log_{10}(SN+1) - .20 + \log_{10} \left( \frac{\Delta PSI}{4.2 - 1.5} \right) \frac{5.19}{(CBR^{.5} + 1)} - .32 \] But for probability, we use the reliability concept: \[ Z_R = \frac{\log_{10}(W_{18}) - X}{S_o} \] Where $X$ is the design value from the AASHTO equation with $Z_R = $ (mean). --- ### **Find $X$ (mean design log ESALs):** #### **Subgrade Resilient Modulus Estimate** For CBR = 8: \[ MR = 150 \times CBR = 150 \times 8 = 12,000 \ \text{psi} \] #### **Plug into AASHTO Equation:** \[ \log_{10}(W_{18}) = 9.36 \log_{10}(SN+1) - .20 + \log_{10} \left( \frac{\Delta PSI}{4.2 - 1.5} \right) + \frac{5.19}{(MR/100 + 1)} - .32 \] - $SN = 4.26$ - $\Delta PSI = P_ - P_t = 4.2 - 2.5 = 1.7$ - $MR = 12,000$ psi Calculate each term: - $\log_{10}(SN+1) = \log_{10}(4.26 + 1) = \log_{10}(5.26) \approx .721$ - $\log_{10}(\Delta PSI / 2.7) = \log_{10}(1.7 / 2.7) = \log_{10}(.63) \approx -.201$ - $MR/100 + 1 = 12 + 1 = 13$ - $5.19/13 \approx .399$ Putting together: \[ \log_{10}(W_{18}) = 9.36 \times .721 - .20 - .201 + .399 - .32 \] \[ = 6.75 - .20 - .201 + .399 - .32 \] \[ = 6.75 - .20 = 6.55 \] \[ 6.55 - .201 = 6.349 \] \[ 6.349 + .399 = 6.748 \] \[ 6.748 - .32 = 6.428 \] So, $X = 6.43$ --- ### **Calculate $Z_R$ (Standard Normal Variable)** \[ W_{18} = 19,200,000 \] \[ \log_{10}(W_{18}) = \log_{10}(19,200,000) \approx 7.28 \] \[ Z_R = \frac{\log_{10}(W_{18}) - X}{S_o} = \frac{7.28 - 6.43}{.45} = \frac{.85}{.45} \approx 1.89 \] --- ## **4. Find Probability from Standard Normal Table** $Z_R = 1.89$ The probability that $Z$ is **less than 1.89** is about **.9706** (from Z-table). That means there is a **97.1% chance** the pavement will NOT exceed this ESALs before reaching PSI 2.5. But the question asks: "**Probability that PSI is above 2.5 after 25 years**" (i.e., pavement lasts 25 years or more before reaching terminal serviceability). So, **Probability that PSI > 2.5 after 25 years = 2.94%** - $P(Z > 1.89) = 1 - .9706 = .0294 = 2.94\%$ --- ## **## Final Answer** ### **Probability that this pavement will have a PSI above 2.5 after 25 years is approximately:** \[ \boxed{2.9\%} \] --- ### **Summary Table** | Step | Value | |---------------------|--------------------------| | SN | 4.26 | | 25-yr ESALs | 19,200,000 | | Design log ESAL ($X$)| 6.43 | | Actual log ESAL | 7.28 | | Z-score ($Z_R$) | 1.89 | | Probability | 2.9% | --- **Note:** If you want the probability that the pavement will fall **below** PSI 2.5 (i.e., fail) in 25 years, it is $97.1\%$. If you want the probability it survives (stays above 2.5), it is $2.9\%$.