--- title: "AASHTO layer thicknesses D1, D2, D3" subject: "Transportation Engineering II" chapter: 2 key: "semester_7/transportation_engineering_ii:aashto-thickness-d1-d2-d3-esal" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/aashto-thickness-d1-d2-d3-esal site: BCT Engineering --- # AASHTO layer thicknesses D1, D2, D3 Transportation Engineering II · Chapter 2: Highway Pavement - Asked in: 2071 Chaitra, 8 marks - Video: [How to find Flexible Pavement thicknesses using AASHTO Method (Structural Number SN calculation)](https://www.youtube.com/watch?v=EvyeFhVWUq0) — Shahid Ullah Khan - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/aashto-thickness-d1-d2-d3-esal - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2 ## Question In the figure below, a pavement system with the resilient moduli, layer coefficient of surface course and drainage coefficients are shown. If predicted ESAL = 15×10⁶, R = 90%, S0 = 0.4, present serviceability index = 4.2 and terminal serviceability index = 2.7, select the thickness of D1, D2 and D3. [Figure: three layers with E1 = 450,000 psi, a1 = 0.44, thickness D1; E2 = 25,000 psi, m2 = 1.2, thickness D2; E3 = 15,000 psi, m3 = 1.2, thickness D3; subgrade MR = 5000 psi] ## Short answer Using AASHTO 1993 with R=90%, S₀=0.4, ΔPSI=1.5, calculated required SN₁=3.42, SN₂=4.18, SN₃=6.08. Thicknesses: D₁ = 8 in (surface), D₂ = 5 in (base), D₃ = 15 in (subbase); total 28 in. ## Answer #### Given data $W_{18}=15\times10^6$, $R=90\%$ ($Z_R=-1.282$), $S_0=0.4$, $\Delta PSI=4.2-2.7=1.5$; layer 1: $E_1=450\,000$ psi, $a_1=0.44$; layer 2: $E_2=25\,000$ psi, $m_2=1.2$; layer 3: $E_3=15\,000$ psi, $m_3=1.2$; subgrade $M_R=5000$ psi. (AASHTO 1993 flexible design.) #### Step 1 - Layer coefficients of layers 2 and 3 $$ a_2=0.249\log_{10}(25000)-0.977=0.118,\qquad a_3=0.227\log_{10}(15000)-0.839=0.109 $$ #### Step 2 - Required structural numbers (from the AASHTO equation, solved for SN) | Support below the layer | $M_R$ (psi) | Required SN | |---|---|---| | Layer 2 (protects with $E_2$) | 25 000 | $SN_1=3.42$ | | Layer 3 (protects with $E_3$) | 15 000 | $SN_2=4.18$ | | Subgrade | 5000 | $SN_3=6.08$ | #### Step 3 - Thicknesses $$ D_1^*\ge\frac{SN_1}{a_1}=\frac{3.42}{0.44}=7.78\ \text{in}\ \Rightarrow\ D_1=8.0\ \text{in},\quad SN_1^*=0.44\times8.0=3.52 $$ $$ D_2^*\ge\frac{SN_2-SN_1^*}{a_2m_2}=\frac{4.18-3.52}{0.118\times1.2}=4.65\ \text{in}\ \Rightarrow\ D_2=5\ \text{in},\quad SN_2^*=0.118\times1.2\times5=0.71 $$ $$ D_3^*\ge\frac{SN_3-(SN_1^*+SN_2^*)}{a_3m_3}=\frac{6.08-(3.52+0.71)}{0.109\times1.2}=14.13\ \text{in}\ \Rightarrow\ D_3=15\ \text{in} $$ Check: $SN=3.52+0.71+1.96=6.19\ge SN_3=6.08$ - safe. **Answer:** $D_1=8.0$ in (20 cm) of surface course, $D_2=5$ in (13 cm) and $D_3=15$ in (38 cm); total 28.0 in. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the design serviceability loss Delta PSI obtained here? A. 4.2 + 2.7 = 6.9 B. 4.2 - 2.7 = 1.5 C. 4.2 / 2.7 = 1.56 D. 4.2 x 0.4 = 1.68 Answer: B. Delta PSI is present serviceability minus terminal serviceability: 4.2 - 2.7 = 1.5. (Key point: Delta PSI) 2. How are SN1, SN2 and SN3 obtained? A. By multiplying a_i by D_i for each layer B. By adding the drainage coefficients C. From the AASHTO equation using the resilient modulus of the layer below each level, with the subgrade giving the largest SN D. From the layer coefficients alone without any design traffic Answer: C. The AASHTO equation is solved for SN with the support modulus below each layer, so SN rises as the support gets weaker (3.42, 4.18, 6.08). (Key point: Required SN at each level) 3. What is the formula for D2? A. D2 >= SN2 / a2 B. D2 >= (SN2 - SN1*) / a2 C. D2 >= (SN3 - SN1*) / (a2 m2) D. D2 >= (SN2 - SN1*) / (a2 m2) Answer: D. The base must supply the SN deficit (SN2 minus the SN of the rounded-up layer 1), and its contribution is a2 x m2 x D2, so the drainage coefficient is included. (Key point: Layer thickness with drainage) 4. Why are D1 = 7.78 in, D2 = 4.65 in and D3 = 14.13 in rounded up to 8, 5 and 15 in? A. Rounding up keeps the provided SN at or above the required SN (6.19 >= 6.08) and gives practical thickness B. Rounding is needed to reduce the cost of materials C. Rounding up makes the layer coefficients larger D. The AASHTO method only permits whole inches Answer: A. Computed values are minimums; rounding up to practical sizes keeps the structure safe, as confirmed by the check SN = 6.19 >= 6.08. 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