Question

Q3: A flexible pavement is to be designed according to the AASHTO design method. The total expected ESAL for a new highway = 9 x10®, APSI= 2, reliability = 99%, overall standard deviation = 0.4, if the pavement layers are: Layer Thickness (inch) Structural layer coefficient Wearing surface 4 ai = 0.44 Base 4 a2 = 0.14 Subbase ? az = 0.11 The subgrade resilient modulus=20,000 psi, m2, mz=1, determine the thickness required for the subbase layer?

Answer 1

[We need to determine the value of SN

The basic equation of flexible pavement design as per AASHTO is given by,

10810 log10 (W18) = {ZR XS, +9.36 x log10 (SN + 1) - 0.20 + APSI 2.7 1094 +2.32 x log10 MR - 8.07 (SN + 1)5.19 0.4 +

Where,

W₁₈ = ESALs

Z_R = Standard normal variate (obtained from standard normal table corresponding to reliability)

S₀ = Overall standard deviation

SN = structural number

ΔPSI = loss in serviceability (change in initial and terminal serviceability index)

ΔPSI = p₀ - p_t

p_t= terminal serviceability index

p_{0 =} initial serviceability index

MR = soil resilient modulus

W₁₈ =9,000,000 (given)

Z_R = –2.326 (obtained from standard normal table corresponding to reliability, R = 99% (given))

Overall standard deviation = S₀ = 0.4 (given)

ΔPSI = 2 (given)

soil resilient modulus (MR)=20,000 psi (given)

soil resilient modulus (MR) = 20,000 psi

Substituting the above values in the equation,

log 10 log10 (W13) = {ZR X So + 9.36 log10 (SN +1) – 0.20 +- ****** APSI 2.7 +2.32 1094 (SN + 1) 5.19 0.4 + X log10 MR - 8.07

log10 (9,000,000) log10 (2) -2.326 X 0.4 +9.36 x log10 (SN +1) - 0.20+. 0.4 + + 2.32 1094 (SN + 1) 5.19 x log10 (20,000) - 8.07

6.95424 = 0.77799 + 9.36 x log10 (SN + 1) +- log10 (2) 1094 0.4 + (SN + 1) 5.19

SN = 3.775

Thus SN = 3.775

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The pavement structural number is determined using the equation,

SN= a_D1 + a2D2M2 + azD3M,

Where,

SN = structural number

'

a; = layer coefficient corresponding to layer 'i'

D= Depth of layer'i

M; = Drainage coefficient corresponding to layer 'i'

Data given in the question are as follows:

Layer 1 refers to wearing surface with Layer coefficient a = 0.44 and

Thickness = D1 = 4"
  

Layer 2 refers to base with Layer coefficient az = 0.14 and

Thickness = D2 = 4"and

Drainage coefficient = M = 1

Layer 3 refers to subbase with Layer coefficient, = az = 0.11 and

Drainage coefficient, M3 = 1

The thickness of subbase is to be determined (D3)

SN = 3.775 (calculated above)

Substituting the above values in the equation,

SN= a_D1 + a2D2M2 + azD3M,

3.775 0.44 X 4 + 0.14 X 4 X 1 + 0.11 X D3 X 1

D3 = 13.227 inches

Thus the thickness of the subbase layer = 13.227"

(safe)

Rounding to the nearest 0.5 inches, we get the thickness of subbase layer = 13.5"

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Answer: The thickness of the subbase layer = 13.5"