How is the structural number of this section obtained?

SN = a1 D1 + a2 D2 m2 using the layer coefficients and thicknesses in inches

SN = D1 + D2 without any coefficients

To find a2, which value of E2 is substituted in a2 = 0.249 log(E2) - 0.977?

30 000 psi (30 ksi converted to psi)

Which value is used as M_R in the AASHTO equation?

The subgrade modulus E3 = 9 ksi = 9000 psi

The surface modulus 365 000 psi

With Z_R = -1.282 and S0 = 0.5, what does the reliability term do, and what is the result?

Subtracts 0.641 from log W18; final capacity about 1.16 x 10^6 ESAL

Adds +0.641 to log W18, giving about 10^7 ESAL

Multiplies SN by 0.5; capacity about 3.3 ESAL

Has no effect; capacity is 10^9 ESAL

2080 Baisakh · 8 marks

AASHTO: ESALs a pavement can carry

Short answer

The base layer coefficient a2 = 0.138 from a2=0.249log(E2)-0.977. Structural number SN = a1D1 + a2D2 = 3.310. Substituting into the AASHTO design equation with MR=9000 psi yields logW18 = 6.066, so the pavement capacity is approximately 1.16 million ESALs.

Figure below shows the section of a flexible asphalt pavement designed using AASHTO method. Determine the number of equivalent single-axle loads (ESALs) that the asphalt pavement structure is capable of carrying at a reliability level (R) of 90% [Z-value is equal to -1.282] overall standard deviation (S0) of 0.5, ΔPSI of 2.0, and m2 = 1.0. Assume material coefficient a1 = 0.401 and coefficient of Layer 2 (the base layer) given by the following formula a2 = 0.249 log(E2) − 0.977

\log_{10} W_{18} = Z_R S_0 + 9.36\log_{10}(SN+1) - 0.20 + \frac{\log_{10}\left[\frac{\Delta PSI}{4.2-1.5}\right]}{0.4+\frac{1094}{(SN+1)^{5.19}}} + 2.32\log_{10} M_R - 8.07

[Figure: HMA surface of thickness 5.3 in with E1 = 365 ksi on a base of thickness 8.6 in with E2 = 30 ksi, over subgrade with E3 = 9 ksi]

Given data

Surface $D_1=5.3$ in, $a_1=0.401$ ; base $D_2=8.6$ in, $E_2=30$ ksi, $m_2=1$ ; subgrade $M_R=E_3=9$ ksi $=9000$ psi; $R=90\%$ , $Z_R=-1.282$ , $S_0=0.5$ , $\Delta PSI=2.0$ .

Step 1 - Layer coefficient of base

a_2=0.249\log_{10}(E_2)-0.977=0.249\log_{10}(30000)-0.977=0.138

Step 2 - Structural number

SN=a_1D_1+a_2D_2m_2=0.401\times5.3+0.138\times8.6\times1.0=2.125+1.185=3.310

Step 3 - ESALs from the AASHTO equation

\begin{aligned} \log_{10}W_{18}&=Z_RS_0+9.36\log_{10}(SN+1)-0.20+\frac{\log_{10}\left[\frac{\Delta PSI}{4.2-1.5}\right]}{0.4+\frac{1094}{(SN+1)^{5.19}}}+2.32\log_{10}M_R-8.07\\ &=(-1.282)(0.5)+9.36\log_{10}(4.310)-0.20+\frac{\log_{10}(2.0/2.7)}{0.4+1094/(4.310)^{5.19}}+2.32\log_{10}(9000)-8.07 \end{aligned}

Terms: $Z_RS_0=-0.641$ ; $9.36\log_{10}(SN+1)=5.939$ ; fatigue term $=-0.136$ ; $2.32\log_{10}M_R=9.174$ .

\log_{10}W_{18}=6.066\ \Rightarrow\ W_{18}=10^{6.066}=1163446

Answer: the pavement can carry about 1.16 $\times10^6$ ESALs (1.16 million) at 90 percent reliability.

✓ Answer checked by Claude Opus 5.5 (high effort)

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AASHTO: ESALs a pavement can carry

Given data

Step 1 - Layer coefficient of base

Step 2 - Structural number

Step 3 - ESALs from the AASHTO equation

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