Chapter 20, Problem 20P

To determine

A suitable thickness of the concrete pavement if the working stress of the concrete is $650\text{lb}/{\text{in}}^{\text{2}}$.

Answer to Problem 20P

The suitable depth of concrete pavement is $7\text{inches}$.

Explanation of Solution

Given:

Design life is $20$ years.

Annual rate of traffic growth is $3.5\%$.

Sub grade $k$ value is $50\text{lb}/{\text{in}}^{\text{2}}$.

Stabilized sub-base is of $6\text{in}$.

Modulus of rupture of the concrete is $600\text{lb}/{\text{in}}^{\text{2}}$.

  $\begin{array}{l}{P}_i=4.5\\ P_t=2.5\\ S_{\text{c}}{}^{\prime }=650\text{lb}/{\text{in}}^{\text{2}}\\ E_{\text{c}}=5\times {10}^6\text{lb}/{\text{in}}^{\text{2}}\end{array}$

  $\begin{array}{l}\text{J}=3.2\\ C_{\text{d}}=1.0\\ S_{\text{a}}=0.3\\ R=9.5\%\end{array}$

Traffic volume data on the highway indicate that the AADT during the first year of operation is $24,000$ with the following vehicle mix and axle loads.

  $\text{Passenger cars}=50\text{ percent}$

  $\text{2-axle single-unit trucks }\left(12,000\text{lb}/\text{axle}\right)=40\text{ percent}$

  $\text{3-axle single-unit trucks }\left(16,000\text{lb}/\text{axle}\right)=10\text{ percent}$

The pavement has aggregate interlock joints (no dowels) and a concrete shoulder.

Formula used:

The design equivalent single axle load is given by,

  $\text{ESAL}=G_{\text{jt}}\times {\text{AADT}}_{\text{i}}\times 365\times N_{\text{i}}\times F_{\text{Ei}}$ ....... (I)

Here, $G_{\text{jt}}$ is the growth factor for a given growth rate $r$ and design period $n$, AADT is the first annual average daily traffic for axle category $i$, $N_{\text{i}}$ is the number of axles on each vehicle in category $i$ and $F_{\text{Ei}}$ is the load equivalency factor for axle category $i$.

The total ESAL is given by,

  $\text{ESAL}={\text{ESAL}}_1+{\text{ESAL}}_2$ ....... (II)

Here, ${\text{ESAL}}_1$ is the equivalent single axle load for $\text{2-axle single-unit trucks}$ and ${\text{ESAL}}_2$ is the $\text{3-axle single-unit trucks}$.

The stress ratio is given by,

  $\text{Stress}\text{ratio}=\frac{\text{Equivalent}\text{stress}}{{S^{\prime }}_{\text{c}}}$ ....... (III)

Here, ${S^{\prime }}_{\text{c}}$ is the working stress of the concrete.

The damage percent is given by,

  $\text{Damage}\text{percent}=\frac{ESA{L}_2}{N}$ ...... (IV)

Here, $N$ is the number of repetitions of load.

Calculation:

The design equivalent single axle load for $\text{2-axle single-unit trucks}$ is calculated as,

Substitute $28.28$ for $G_{\text{jt}}$, $24,000$ for AADT, $2$ for $N_{\text{i}}$ and $0.40$ for $F_{\text{Ei}}$ in equation (I).

  $\begin{array}{c}{\text{ESAL}}_1=28.28\times 24,000\times 365\times 2\times 0.40\\ =79.27\times {10}^6\end{array}$

The design equivalent single axle load for $\text{3-axle single-unit trucks}$ is calculated as,

Substitute $28.28$ for $G_{\text{jt}}$, $24,000$ for AADT, $3$ for $N_{\text{i}}$ and $0.40$ for $F_{\text{Ei}}$ in equation (I).

  $\begin{array}{c}{\text{ESAL}}_2=28.28\times 24,000\times 365\times 3\times 0.40\\ =29.73\times {10}^6\end{array}$

The design headwater depth is calculated as,

Substitute $79.27\times {10}^6$ for ${\text{ESAL}}_1$ and $29.73\times {10}^6$ for ${\text{ESAL}}_2$ in equation (II).

  $\begin{array}{c}\text{ESAL}=79.27\times {10}^6+29.73\times {10}^6\\ =109\times {10}^6\end{array}$

The sub grade value of $k$ is $167\text{lb}/{\text{in}}^{\text{3}}$.

From table 20.22, "Deign $k$ values for Untreated and Cement-Treated Sub bases" of book "Traffic and highway engineering" equivalent value of $k$ can be obtained by interpolation.

  $\begin{array}{c}k=140+\left(\frac{230-140}{100}\right)\times \left(130-100\right)\\ =140+27\\ =167\text{lb}/{\text{in}}^{\text{3}}\end{array}$

Assume a slab $6\text{in}$ thick with doweled joints and concrete shoulders.

From table 20.24, "Equivalent stress values for single axles and tandem axles" of book "Traffic and highway engineering" equivalent stress values can be obtained and for $6.5\text{inches}$ thick slab by interpolation effective $k$ can be obtained as,

  $\begin{array}{c}\text{Equivalent}\text{stress}=304-\left[\left(304-289\right)\times \frac{\left(200-167\right)}{\left(200-150\right)}\right]\\ =304-9.9\\ =294.1\text{lb}/{\text{in}}^{\text{2}}\end{array}$

The stress ratio is calculated as,

Substitute $294.1\text{lb}/{\text{in}}^{\text{3}}$ for equivalent stress and $650\text{lb}/{\text{in}}^{\text{2}}$ for ${S^{\prime }}_{\text{c}}$ in equation (III).

  $\begin{array}{c}\text{Stress}\text{ratio}=\frac{294.1\text{lb}/{\text{in}}^{\text{2}}}{650\text{lb}/{\text{in}}^{\text{2}}}\\ =0.45\end{array}$

From table 20.27, "Erosion factors for single axles and tandem axles (doweled joint, concrete shoulder" of book "Traffic and highway engineering" for $6.5\text{inches}$ slab the erosion factor is $2.72$.

  $\begin{array}{l}12,000\left(1.2\right)=14,400\text{lb}/\text{axle}\\ 16,000\left(1.2\right)=19,200\text{lb}/\text{axle}\end{array}$

From figure 20.26, "Allowable Load Repetitions for Fatigue Analysis Based on Stress Ratio" of book "Traffic and highway engineering" for $14.4\text{kip}$ single axle load to $0.45$ stress ratio factor the allowable load repetitions for fatigue analysis is unlimited.

Thus, the Fatigue analysis for $14,400\text{lb}/\text{axle}$ is unlimited.

From figure 20.27, "Allowable Load Repetitions for Erosion Analysis Based on erosion factors" of book "Traffic and highway engineering" for $14.4\text{kip}$ single axle load to $2.72$ erosion factor the allowable load repetitions for erosion analysis is unlimited.

From figure 20.26, "Allowable Load Repetitions for Fatigue Analysis Based on Stress Ratio" of book "Traffic and highway engineering" for $19.2\text{kip}$ single axle load to $0.45$ stress ratio factor the allowable load repetitions for fatigue analysis is $7\times {10}^5$.

From figure 20.27, "Allowable Load Repetitions for Erosion Analysis Based on erosion factors" of book "Traffic and highway engineering" for $14.4\text{kip}$ single axle load to $2.72$ erosion factor the allowable load repetitions for erosion analysis is $2.6\times {10}^6$.

The table showing the allowable repetitions for $6\text{inches}$ slab is given below.

Load	Fatigue analysis	Erosion analysis
$14,400\text{lb}/\text{axle}$	Unlimited	Unlimited
$19,200\text{lb}/\text{axle}$	$7\times {10}^5$	$2.6\times {10}^6$

Table (1)

So the $6\text{inches}$ slab is inadequate since the allowable ESALs are less than the expected ESALs.

Assume a slab $7\text{in}$ thick with doweled joints and concrete shoulders.

Calculate the equivalent stress.

From table 20.24, "Equivalent stress values for single axles and tandem axles" of book "Traffic and highway engineering" equivalent stress values can be obtained and for $6.5\text{inches}$ thick slab by interpolation effective $k$ can be obtained as,

  $\begin{array}{c}\text{Equivalent}\text{stress}=248-\left[\left(248-236\right)\times \frac{\left(200-167\right)}{\left(200-150\right)}\right]\\ =248-7.92\\ =240.08\text{lb}/{\text{in}}^{\text{2}}\end{array}$

The stress ratio is calculated as,

Substitute $240.08\text{lb}/{\text{in}}^{\text{2}}$ for equivalent stress and $650\text{lb}/{\text{in}}^{\text{2}}$ for ${S^{\prime }}_{\text{c}}$ in equation (III).

  $\begin{array}{c}\text{Stress}\text{ratio}=\frac{240.08}{650}\\ =0.37\end{array}$

From table 20.27, "Erosion factors for single axles and tandem axles (doweled joint, concrete shoulder" of book "Traffic and highway engineering" for $7\text{inches}$ slab the erosion factor is $2.54$.

  $\begin{array}{l}12,000\left(1.2\right)=14,400\text{lb}/\text{axle}\\ 16,000\left(1.2\right)=19,200\text{lb}/\text{axle}\end{array}$

From figure 20.26, "Allowable Load Repetitions for Fatigue Analysis Based on Stress Ratio" of book "Traffic and highway engineering" for $14.4\text{kip}$ single axle load to $0.37$ stress ratio factor the allowable load repetitions for fatigue analysis is unlimited.

Thus, the Fatigue analysis for $14,400\text{lb}/\text{axle}$ is unlimited.

From figure 20.27, "Allowable Load Repetitions for Erosion Analysis Based on erosion factors" of book "Traffic and highway engineering" for $14.4\text{kip}$ single axle load to $2.54$ erosion factor the allowable load repetitions for erosion analysis is unlimited.

From figure 20.26, "Allowable Load Repetitions for Fatigue Analysis Based on Stress Ratio" of book "Traffic and highway engineering" for $19.2\text{kip}$ single axle load to $0.37$ stress ratio factor the allowable load repetitions for fatigue analysis is unlimited.

From figure 20.27, "Allowable Load Repetitions for Erosion Analysis Based on erosion factors" of book "Traffic and highway engineering" for $19.2\text{kip}$ single axle load to $2.54$ erosion factor the allowable load repetitions for erosion analysis is $100\times {10}^6$.

The table showing the allowable repetitions for $6\text{inches}$ slab is given below.

Load	Fatigue analysis	Erosion analysis
$14,400\text{lb}/\text{axle}$	Unlimited	Unlimited
$19,200\text{lb}/\text{axle}$	unlimited	$100\times {10}^6$

Therefore, the $7\text{in}$ slab is adequate. Use a $7\text{in}$ slab depth.

Conclusion:

Therefore, the suitable depth of concrete pavement is $7\text{inches}$.