Chapter 19, Problem 13P

Repeat Problem 19-7 using two different depths of untreated aggregate bases of 6 in. and 12 in. Highway contractors in your area can furnish rates for providing and properly laying an asphalt concrete surface and untreated granular base. Assume a structural coefficient of 0.12 for the base course. If these rates are available, determine the cost for constructing the different pavement designs if the highway section is 5 miles long and the lane width is 12 ft. Which design will you select for construction?

To determine

The cost for constructing the different pavement designs.

Answer to Problem 13P

  $\underset{\_}{\text{\$ 693792}\text{.}}$

Explanation of Solution

Given information:

Following is the given information:

Equivalent single axle load, ESAL = $0.4354\times {10}^6.$,

CBR = $85$.

Subgrade resilient modulus = $18,000{\text{lb/in}}^{\text{2}}$,

Sub-base layer coefficient = $0.13,$

Granular base layer coefficient = $0.14,$

Elastic modulus of asphalt concrete = $280,000{\text{lb/in}}^{\text{2}},$

m_i = 1, Percentage of traffic on design lane = $45\%,$

SN = 4, reliability level = $85\%,$ standard deviation = $0.45,$

and design serviceability loss = $2.0$.

Calculation:

We have the following formula for the calculation of truck factor:

  $\text{ESAL }=f_d\times G_{rn}\times AADT\times 365\times N_i\times F_{Ei}$

Where, ESAL i= equivalent accumulated 18,000-lb (80-kN) single-axle load for the axle category i

f_d= design lane factor,

G _rn= growth factor for a given growth rate r and design period n

AADT i= first-year annual average daily traffic for axle category i

N _i= number of axles on each vehicle in category i

F _Ei= load equivalency factor for axle category i

Calculate ESAL for passenger car$2000\text{lb/axle}$ single unit:

  $\begin{array}{l}\text{ESAL }=f_d\times G_{rn}\times AADT\times 365\times N_i\times F_{Ei}\\ \text{ESAL }=0.45\times 33.06\times 500\times 365\times 2\times 0.0002\\ \text{ESAL }=1086.021\\ \text{ESAL }=0.01086021\times {10}^6\end{array}$

We have the following formula for the calculation of design serviceability :

  $\Delta =P_i-P_t$

Substitute the values in the required equation.

Initial serviceability index, $P_i=4.5$

Terminal serviceability index, $P_t=2.5$

  $\begin{array}{l}\Delta =4.5-2.5\\ \Delta =2.0\end{array}$

Let's determine the resilient modulus of subgrade:

The resilient modulus of subgrade is 1500 times CBR

The value of resilient modulus is given as follows:

  $\begin{array}{l}{M}_r=85\times 1500{\text{lb/in}}^{\text{2}}.\\ M_r=127500{\text{lb/in}}^{\text{2}}.\end{array}$

Now for the structural number, SN refer to figure 19.10

The reliability level is 90%. Starting from the point 90% in figure, extend this point to standard deviation 0.45. From standard deviation point, extend this line to line A as given in the figure.

From line A extend this line to $0.001\times {10}^6\text{ESAL}$

From ESAL extend this line to line B in the figure.

From line B extend this line to $127500{\text{lb/in}}^{\text{2}}\text{in }{\text{M}}_{\text{r}}.$

From M _rextend this line till it touches the graph. From that point, draw a horizontal line to touch the design serviceability loss, which is 2.

Extend a line vertically and take the structural number SN₃reading, which is 2.5 and SN₂is 2.3

Then from figure 19.6, for layer coefficient and the resilient modulus for base layer is

  $\begin{array}{l}{a}_2=0.12\\ E_s=25000{\text{lb/in}}^{\text{2}}\end{array}$

From figure 19.10, resilient modulus is $25000{\text{lb/in}}^{\text{2}}$. In structural number SN₁is 1.6

Refer figure 19.7 to determine layer coefficient $a_1$, from figure 19.7 for resilient modulus of asphalt $E_{AC}=400,000{\text{lb/in}}^{\text{2}},a_1=0.43$

Now, considering the thickness aggregate base as 6 inches.

Calculate the minimum thickness.

  $D_1=\frac{S{N}_1}{a_1}$

The layer coefficient, $a_2=0.14$ from figure 19.6 and $a_1=0.43$ from Figure 19.7 Chart for Estimating Structural Layer Coefficient of Dense-Graded/Asphalt Concrete

  $S{N}_1=1.6$ from the figure 19.10

Substitute the values

  $\begin{array}{l}{D}_1=\frac{S{N}_1}{a_1}\\ D_1=\frac{1.6}{0.43}\\ D_1=3.7209\\ D_1\approx 4\text{inch}\text{.}\end{array}$

Checking the values, we have

  $\begin{array}{l}S{N}_1=a_1{D}_1\\ S{N}_1=0.43\times 4\\ S{N}_1=1.72\end{array}$

As the values match, thus the thickness of subgrade calculated is correct.

Calculate the required minimum thickness of base course D₂is 6 inches.

Check whether the calculated thickness agrees with the calculated structural number using equation

  ${\text{SN}}_{\text{2}}\text{=}{\text{a}}_{\text{2}}{\text{D}}_{\text{2}}{\text{m}}_{\text{2}}{\text{+SN}}_{\text{1}}$

Substituting the values, we have

  $\begin{array}{l}{\text{SN}}_{\text{2}}\text{=}\left(\text{0}\text{.12}\times \text{6}\times \text{1}\right)\text{+ 1}\text{.7}\\ {\text{SN}}_{\text{2}}\text{=}\text{0}\text{.72}\text{+}\text{1}\text{.7}\\ {\text{SN}}_{\text{2}}\text{=}2.42\end{array}$

Which is compatible with the structural number calculated already.

And implies that the thickness of the surface calculated is correct.

Now, the required minimum thickness of the sub-base course $D_3$ is

  $D_3\ge \frac{S{N}_3-S{N}_2}{a_3{m}_3}$

By substituting the values, we have

  $\begin{array}{l}{D}_3\ge \frac{2.5-2.42}{0.13\times 1}\\ D_3\ge \frac{0.08}{0.13}\\ D_3\ge 0.615\text{lets take 1}\text{.00 inch}\end{array}$

Now, check if the calculated thickness agrees with the calculated structural number of using the following equation:

  ${\text{SN}}_3\text{=}{\text{a}}_3{\text{D}}_3{\text{m}}_3\text{+}{\text{SN}}_2\text{+}{\text{SN}}_{\text{1}}$

Substituting the values

  $\begin{array}{l}{\text{SN}}_3\text{=}\left(\text{0}\text{.13}\times 1\times \text{1}\right)\text{+ 2}\text{.42}\text{+1}\text{.7}\\ {\text{SN}}_3\text{=}\text{0}\text{.13}\text{+}4.12\\ {\text{SN}}_3\text{=}4.25\end{array}$

Which is compatible with the structural number calculated already.

And implies that the thickness of the surface calculated is correct.

The thickness of surface 4.25 inches.

Now, considering the thickness aggregate base as 12 inches.

Calculate the minimum thickness.

  $D_1=\frac{S{N}_1}{a_1}$

Substitute the values, we have

  $\begin{array}{l}{D}_1=\frac{S{N}_1}{a_1}\\ S{N}_1=D_1\times a_1\\ S{N}_1=0.43\times 4\\ S{N}_1\approx 1.7\text{inch}\text{.}\end{array}$

Thus, the thickness of sub grade is 1.7 in.

Now, the thickness of the base course is given as follows:

  ${\text{SN}}_2\text{=}{\text{a}}_2{\text{D}}_2{\text{m}}_2\text{+}{\text{SN}}_1$

Substitute the values.

  $\begin{array}{l}{\text{SN}}_2\text{=}{\text{a}}_2{\text{D}}_2{\text{m}}_2\text{+}{\text{SN}}_1\\ {\text{SN}}_2\text{=}\left(0.12\times 12\times 1\right)\text{+}\left(1.7\right)\\ {\text{SN}}_2\text{=}\left(0.12\times 12\right)\text{+}\left(1.7\right)\\ {\text{SN}}_2\text{=}\left(1.44\right)\text{+}\left(2.57\right)\\ {\text{SN}}_2\text{=}4.01\end{array}$

Which is greater than 2.7, i.e., structural number.

Now, the required minimum thickness of the sub-base course $D_3$ is

  $D_3\ge \frac{S{N}_3-S{N}_2}{a_3{m}_3}$

By substituting the values,

  $\begin{array}{l}{D}_3\ge \frac{2.5-4.01}{0.13\times 1}\\ D_3\ge \frac{-1.51}{0.13}\\ D_3\ge -11.62\text{lets take 12 inch}\end{array}$

Now, check if the calculated thickness agrees with the calculated structural number of using the following equation

  ${\text{SN}}_3\text{=}{\text{a}}_3{\text{D}}_3{\text{m}}_3\text{+}{\text{SN}}_2\text{+}{\text{SN}}_{\text{1}}$

Substituting the values,

  $\begin{array}{l}{\text{SN}}_3\text{=}\left(\text{0}\text{.13}\times -12\times \text{1}\right)\text{+ 4}\text{.01}\text{+1}\text{.7}\\ {\text{SN}}_3\text{=}-1.56\text{+}5.71\\ {\text{SN}}_3\text{=}4.15\end{array}$

Which is compatible with the structural number calculated already.

And implies that the thickness of the surface calculated is correct.

Calculate construction cost for the highway considering thickness of base layer as 6 inches and 12 inches.

The cost is calculated by multiplying the measurement of road with the cost per rate.

Assume cost of surface layer as $D_1=\$1/\text{ft}.$

For cost of base layer $D_2=\$2/\text{ft}.$

For cost of sub-base layer $D_3=\$2/\text{ft}.$

The construction cost of the highway is as follows:

Cost = L X B X t X C

For surface layer :

Substitute the values

5 miles is equal to 26400

  $\begin{array}{l}\text{Cost = L}\times \text{ B }\times \text{ t}\times \text{ C}\\ \text{Cost = 26400}\times \text{ 12}\times \text{ 0}\text{.17}\times \text{ 1}\\ \text{Cost = \$ 53856}\text{.}\end{array}$

For base layer :

Substitute the values

  $\begin{array}{l}\text{Cost = L}\times \text{ B }\times \text{ t}\times \text{ C}\\ \text{Cost = 26400}\times \text{ 12}\times \text{ 0}\text{.5}\times \text{2}\\ \text{Cost = \$ 316800}\text{.}\end{array}$

For sub-base layer :

Substitute the values

  $\begin{array}{l}\text{Cost = L}\times \text{ B }\times \text{ t}\times \text{ C}\\ \text{Cost = 26400}\times \text{ 12}\times \text{ 0}\text{.34}\times \text{3}\\ \text{Cost = \$ 323136}\text{.}\end{array}$

Now, calculate the total cost as follows:

  $\begin{array}{l}\text{Total cost = \$ 53856}+\text{\$ 316800}+\text{\$ 323136}\text{.}\\ \text{Total cost = \$ 693792}\text{.}\end{array}$

The construction cost of the highway is as follows:

Cost = L X B X t X C

For surface layer :

Substitute the values

5 miles is equal to 26400

  $\begin{array}{l}\text{Cost = L}\times \text{ B }\times \text{ t}\times \text{ C}\\ \text{Cost = 26400}\times \text{ 12}\times \text{ 0}\text{.34}\times \text{ 1}\\ \text{Cost = \$ 107712}\text{.}\end{array}$

For base layer :

Substitute the values

  $\begin{array}{l}\text{Cost = L}\times \text{ B }\times \text{ t}\times \text{ C}\\ \text{Cost = 26400}\times \text{ 12}\times \text{ 1}\times \text{2}\\ \text{Cost = \$ 633600}\text{.}\end{array}$

For sub-base layer :

Substitute the values

  $\begin{array}{l}\text{Cost = L}\times \text{ B }\times \text{ t}\times \text{ C}\\ \text{Cost = 26400}\times \text{ 12}\times \text{ 0}\text{.42}\times \text{3}\\ \text{Cost = \$ 399168}\text{.}\end{array}$

Now, calculate the total cost as follows:

  $\begin{array}{l}\text{Total cost = \$ 107712}+\text{\$ 633600}+\text{\$ 399168}\text{.}\\ \text{Total cost = \$ 1140480}\text{.}\end{array}$

Conclusion:

Therefore, out of two cases the construction cost is $\text{\$ 693792}\text{.}$ As the cost in this case is less and the thickness of the layers calculated are more accurate.