How are SN1, SN2 and SN3 obtained?

From the AASHTO equation using the resilient modulus of the layer below each level, with the subgrade giving the largest SN

By multiplying a_i by D_i for each layer

By adding the drainage coefficients

From the layer coefficients alone without any design traffic

Why are D1 = 7.78 in, D2 = 4.65 in and D3 = 14.13 in rounded up to 8, 5 and 15 in?

Rounding up keeps the provided SN at or above the required SN (6.19 >= 6.08) and gives practical thickness

Rounding is needed to reduce the cost of materials

Rounding up makes the layer coefficients larger

The AASHTO method only permits whole inches

2071 Chaitra · 8 marks

AASHTO layer thicknesses D1, D2, D3

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.

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]

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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AASHTO layer thicknesses D1, D2, D3

Given data

Step 1 - Layer coefficients of layers 2 and 3

Step 2 - Required structural numbers (from the AASHTO equation, solved for SN)

Step 3 - Thicknesses

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