Question

(18%) An engineer wants to design a flexible pavement which has a 4-inch sand-mix asphalt wearing surface, 11-inch soil cement base, and a 11-inch crushed stone subbase. It is designed to withstand 400 22-kip single-axle loads and 800 34-kip tandem axle loads per day. The subgrade CBR is 8, the overall standard deviation is 0.45, the initial PSI is 4.2, and the final PSI is 2.5. What is the probability that this pavement will have a PSI above 2.5 after 25 years? (Drainage coefficients are 1.0.) **Round up the SN value to nearest integer**

Answer 1

Thickness of sand - mix asphalt wearing surface (DZ) = 4 inches

Thickness of soil cement base (D2) 11 inches

Thickness of crushed stone sub – base (D3) 11 inches

Drainage coefficient for base = M2 1 (given)

Drainage coefficient for sub-base = M3 = 1 (given)

As per the table for structural layer coefficients as per AASHTO design,

Layer coefficient of sand - mix asphalt wearing surface a = 0.35

Layer coefficient of soil cement base = =a2 = 0.2

Layer coefficient of crushed stone subbase = аз 0.11

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The pavement structural number:

                 

SN a Di +a,D2M2 + a,D2M2

Where,

SN = structural number

a; = layer coefficient corresponding to layer 'i'

Di = Depth of layer'i

M = Drainage coefficient corresponding to layer 'i'

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D. = 4"

D2 = 11"

D3 11"

a = 0.35

az = 0.2

az 3 = 0.11

M2 = 1

M3 = 1

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Substituting the above values in the equation,

SN = a D. + a D2M2 +a, D3M,

SN 0.35 X 4+ 0.2 X 11 X 1 + 0.11 X 11 X 1

SN = 4.81

Thus, SN = 5 (Rounding off to near integer)

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Cumulative equivalent single axle load (AASHTO design)

(single axle

TSI = 2.5 (given)

SN = 5 (calculated above)): Axle load equivalency factor for 22 kips = 2.18

(tandem axle,

TSI = 2.5 (given)

SN = 5 (calculated above)): Axle load equivalency factor for 34 kips = 1.09

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• 18 kip-ESAL or axle load equivalency factor for 400 passes (22 kip single axles)

18 kip - ESAL = 400 x Axle load equivalency factor for 22 kips single axle

18 kip - ESAL = 400 x 2.18

18 kip - ESAL = 872

• 18 kip-ESAL or axle load equivalency factor for 800 passes (34 kip single axles)

18 kip - ESAL = 800 x Axle load equivalency factor for 34 kips single axle

18 kip - ESAL = 800 x 1.09

18 kip - ESAL = 872

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Cumulative equivalent single axle loads for daily traffic =

            18 kip – ESAL for 400 passes (22 kip single axles) +

            18 kip – ESAL for 800 passes (34 kip single axles)

Cumulative equivalent single axle loads for daily traffic = 872+872

Cumulative equivalent single axle loads for daily traffic = 1744

Cumulative equivalent single axle loads over 25 year design period =

Cumulative equivalent single axle loads for daily traffic x 365 x 25

Cumulative equivalent single axle loads over 25 year design period = 1744 x 365 x 25

Cumulative equivalent single axle loads over ‘n’ year design period = 15914000 18 kip – ESAL

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The basic equation of flexible pavement design as per AASHTO

log10 (37) log10 (W13) = ZRX 50 + 9.36 x log10 (SN + 1) - 0.20 +; :+2.32 x log10 MR - 8.07 0.4+ 1094 (SN-195.19

Where,

W₁₈ = Cumulative equivalent single axle loads over ‘n’ year design period (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

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

ΔPSI = p₀ - p_t

p₀ = initial serviceability index = 4.2

p_t= Terminal serviceability index (TSI) = 2.5 (given)

ΔPSI = p₀ - p_t

ΔPSI = 4.2– 2.5

ΔPSI = 1.7

soil resilient modulus (MR) = 1500 X CBR (if CBR less than 10)

Given, CBR = 8

Thus,

soil resilient modulus (MR) = 1500 X CBR

soil resilient modulus (MR) = 1500 x &

soil resilient modulus (MR) = 12000 lb/in?

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MR = 12000 lb/in?

W₁₈ = 15914000

S₀ = 0.45

SN = 5

ΔPSI = 1.7

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Substituting the above values in the equation,

log10 log10 (W13) = = {ZR X S. +9.36 x log10 (SN + 1) -0.20 + 0.4 + (APSI 2.7 1094 + 2.32 x log10 MR - 8.07 (SN + 1) 5.19

log10 (15914000 ) = ZR 0.45 + 9.36 x log10 (5+1) - 0.20 + 0.4 + 10810 (37) - +2.32 x log10 (12,000) - 8.07 1094 6 + 1)5.19

Solving,

ZR = -1.941

Standard normal variate = -1.941

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Reliability (percent probability)

As per standard normal table

ZR = -1.881 with reliability of 97 %

ZR = -2.054 with reliability of 98 %

We know that

ZR = -1.941

Thus by interpolation, we get reliability of 97.346 % corresponding to ZR = -1.941

ZR = -1.941 with reliability of 97.346 %

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Answer:

Probability that the pavement will have PSI above 2.5 after 25 years is 97.346 %