Chapter 4, Problem 1P

To determine

The truck that will cause more pavement damage.

Answer to Problem 1P

The truck A will cause more pavement damage.

Explanation of Solution

The truck A will cause more pavement damage.

Given:

The truck A has two singles axles.

Weight of one axle of truck A is $12,000\text{lb}$ .

Weight of other axle of truck A is $23,000\text{lb}$ .

Weight of single axle of truck B is $8000\text{lb}$ .

Weight of the tandem axle of truck B is $43,000\text{lb}$ .

The thickness of hot mix asphalt is $3\text{inch}$ .

The thickness of soil-cement base is $6\text{inch}$ .

The thickness of crushed stone sub base is $8\text{inch}$ .

The drainage coefficients are $1.0$ .

Formula Used:

Write the expression for the structural number.

  $\text{SN}=a_1{D}_1+a_2{D}_2{M}_2+a_3{D}_3{M}_3$ …… (I)

Here, $a_1$ is the structural layer coefficient of the hot mix asphalt wearing surface layer, $a_2$ is the structural layer coefficient of soil-cement base layer, $a_3$ is the structural coefficient of crushed stone sub-base layer, $D_1$ is the thickness of the wearing surface layer, $D_2$ is the thickness of the base layer, $D_3$ is the thickness of the sub-base layer, $M_2$ is the drainage coefficient for base and $M_3$ is the drainage coefficient for sub-base.

Write the expression for equivalent singleaxle load for truck A.

  ${\text{W}}_{\text{A}}={\text{AF}}_{12}+{\text{AF}}_{23}$ ……. (II)

Here, ${\text{W}}_{\text{A}}$ is the equivalent single axle load factor for truck A, ${\text{AF}}_{\text{12}}$ is the axle load factor for $12\text{kip}$ load and ${\text{AF}}_{\text{23}}$ is the axle load factor for $23\text{kips}$ load.

Write the expression for equivalent single axle load for truck B.

  ${\text{W}}_{\text{B}}={\text{AF}}_8+{\text{AF}}_{43}$ ……. (III)

Here, ${\text{W}}_{\text{B}}$ is the equivalent single axle load factor for truck B, ${\text{AF}}_{\text{8}}$ is the axle load factor for $8\text{kip}$ load and ${\text{AF}}_{\text{43}}$ is the axle load factor for $43\text{kips}$ load.

Calculation:

Refer Table $4.5$ “Structural-layer coefficients” of the book “Principles of Highway Engineering and traffic, $6^{\text{th}}$ Edition”

For the hot-mix asphaltic concrete,

  $a_1=0.44$

For the soil cement base,

  $a_2=0.2$

For the crushed stone,

  $a_3=0.11$

Substitute $0.44$ for $a_1$ , $3$ for $D_1$ , $0.2$ for $a_2$ , $6$ for $D_2$ , $1$ for $M_2$ , $0.11$ for $a_3$ , $8$ for $D_3$ and $1$ for $M_3$ in equation (I).

  $\begin{array}{c}SN=\left(0.44\times 3\right)+\left(0.2\times 6\times 1\right)+\left(0.11\times 8\times 1\right)\\ =1.32+1.2+0.88\\ =3.4\end{array}$

Consider truck A.

Refer Table $4.1$ “Axle-Load Equivalency Factors for Flexible Pavements, Single Axles,and $\text{TSI}=2.5$ ” of the book “Principles of Highway Engineering and traffic, $6^{\text{th}}$ Edition.”

The axle load equivalency factor corresponding to pavement structural number $3$ and axle load $12\text{kips}$ is $0.229$ .

The axle load equivalency factor corresponding to pavement structural number $4$ and axle load $12\text{kips}$ is $0.213$ .

Calculate the axle load equivalency factor corresponding to pavement structural number $3.4$ using interpolation.

  $\begin{array}{c}x=0.229-\left[\frac{\left(3.4-3\right)\left(0.229-0.213\right)}{4-3}\right]\\ =0.229-\left(6.4\times {10}^{-3}\right)\\ =0.2226\end{array}$

The axle load equivalency factor corresponding to pavement structural number $3$ and axle load $22\text{kips}$ is $2.17$ .

The axle load equivalency factor corresponding to pavement structural number $3$ and axle load $24\text{kips}$ is $3.09$ .

The axle load equivalency factor corresponding to pavement structural number $3$ and axle load $23\text{kips}$ is calculated as,

  $\begin{array}{c}\text{Axle}\text{load}\text{factor}=\frac{2.17+3.09}{2}\\ =\frac{5.26}{2}\\ =2.63\end{array}$

The axle load equivalency factor corresponding to pavement structural number $4$ and axle load $22\text{kips}$ is $2.09$ .

The axle load equivalency factor corresponding to pavement structural number

  $4$ and axle load $24\text{kips}$ is

  $2.89$ .

The axle load equivalency factor corresponding to pavement structural number

  $4$ and axle load $23\text{kips}$ is calculated as,

  $\begin{array}{c}\text{Axle}\text{load}\text{factor}=\frac{2.09+2.89}{2}\\ =\frac{4.98}{2}\\ =2.49\end{array}$

Calculate the axle load equivalency factor corresponding to pavement structural number $3.4$ and axle load $23\text{kips}$ using interpolation.

  $\begin{array}{c}x=2.63-\left[\frac{\left(3.4-3\right)\left(2.63-2.49\right)}{4-3}\right]\\ =2.63-\left(0.056\right)\\ =2.574\end{array}$

Substitute $0.2226$ for ${\text{AF}}_{12}$ and $2.574$ for ${\text{AF}}_{\text{23}}$ in equation (II).

  $\begin{array}{c}{\text{W}}_{\text{A}}=0.2226+2.574\\ =2.7966\end{array}$

Consider truck B.

Refer Table $4.1$ “Axle-Load Equivalency Factors for Flexible Pavements, Single Axles, and $\text{TSI}=2.5$ ” of the book “Principles of Highway Engineering and traffic, $6^{\text{th}}$ Edition.”

The axle load equivalency factor corresponding to pavement structural number $3$ and axle load $8\text{kips}$ is $0.051$ .

The axle load equivalency factor corresponding to pavement structural number $4$ and axle load $8\text{kips}$ is $0.041$ .

Calculate the axle load equivalency factor corresponding to pavement structural number $3.4$ using interpolation.

  $\begin{array}{c}x=0.051-\left[\frac{\left(3.4-3\right)\left(0.051-0.041\right)}{4-3}\right]\\ =0.051-\left(4\times {10}^{-3}\right)\\ =0.047\end{array}$

Refer Table $4.2$ “Axle-Load Equivalency Factors for Flexible Pavements, Tandem Axles, and $\text{TSI}=2.5$ ” of the book “Principles of Highway Engineering and traffic, $6^{\text{th}}$ Edition.”

The axle load equivalency factor corresponding to pavement structural number $3$ and axle load $42\text{kips}$ is $2.49$ .

The axle load equivalency factor corresponding to pavement structural number $3$ and axle load $44\text{kips}$ is $2.99$ .

The axle load equivalency factor corresponding to pavement structural number $3$ and axle load $43\text{kips}$ is calculated as,

  $\begin{array}{c}\text{Axle}\text{load}\text{factor}=\frac{2.49+2.99}{2}\\ =\frac{5.48}{2}\\ =2.74\end{array}$

The axle load equivalency factor corresponding to pavement structural number $4$ and axle load $42\text{kips}$ is $2.43$ .

The axle load equivalency factor corresponding to pavement structural number $4$ and axle load $44\text{kips}$ is $2.88$ .

The axle load equivalency factor corresponding to pavement structural number $4$ and axle load $43\text{kips}$ is calculated as,

  $\begin{array}{c}\text{Axle}\text{load}\text{factor}=\frac{2.43+2.88}{2}\\ =\frac{5.31}{2}\\ =2.655\end{array}$

Calculate the axle load equivalency factor corresponding to pavement structural number $3.4$ and axle load $43\text{kips}$ using interpolation.

  $\begin{array}{c}x=2.74-\left[\frac{\left(3.4-3\right)\left(2.74-2.655\right)}{4-3}\right]\\ =2.74-\left(0.034\right)\\ =2.706\end{array}$

Substitute $0.047$ for ${\text{AF}}_8$ and $2.706$ for ${\text{AF}}_{\text{43}}$ in equation (III).

  $\begin{array}{c}{\text{W}}_{\text{B}}=0.047+2.706\\ =2.753\end{array}$

Since, ${\text{W}}_{\text{A}}>{\text{W}}_{\text{B}}$ . Hence, truck A will cause more damage.

Conclusion:

Thus, the truck A will cause more damage.