Chapter 20, Problem 27P

To determine

The expected speed predicted transverse cracking for the associated applied stresses and five traffic levels.

Answer to Problem 27P

  $\text{CRK=0}\text{.309}$

Explanation of Solution

Given information:

Modulus of rupture of the PCC is $\text{=650 lb/i}{\text{n}}^2$.

Calculation:

• Determine the allowable number of load applications for each axle level load

  $\left(\text{use Eq}\text{. 20}\text{.43}\right)$ from textbook "Traffic and Highway Engineering, 5th Edition".

  $\begin{array}{l}\text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=}{\text{C}}_{\text{1}}{\left(\frac{{\text{S}}^{\text{i}}{}_{\text{c,i}}}{{\text{σ}}_{\text{i}\text{.j,k,l,m,n}}}\right)}^{{\text{C}}_{\text{2}}}\mathrm{.......}(1)\\ \text{where}\\ {\text{N}}_{\text{i}\text{.j,k,l,m,n}}\text{ = allowable number of load applications at condition i,j,k,l,m,n}\\ {\text{S}}^{\text{i}}{}_{\text{c,i}}\text{=650=PCC modulus rupture at age i (lb/i}{\text{n}}^{\text{2}}\text{)}\\ {\text{σ}}_{\text{i}\text{.j,k,l,m,n}}\text{=300=applied stress at condition i,j,k,l,m,n}\\ {\text{C}}_{\text{1}}\text{=calibration constant, 2}\text{.0}\\ {\text{C}}_{\text{2}}\text{=calibration constant, 1}\text{.22}\end{array}$

Substitute values in equation $(1)$ we get,

  $\begin{array}{l}\text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=}{\text{C}}_{\text{1}}{\left(\frac{{\text{S}}^{\text{i}}{}_{\text{c,i}}}{{\text{σ}}_{\text{i}\text{.j,k,l,m,n}}}\right)}^{{\text{C}}_{\text{2}}}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=2}{\left(\frac{\text{650}}{300}\right)}^{1.22}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=2}{\left(2.166\right)}^{1.22}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=}5.137\\ {\text{N}}_{\text{i}\text{.j,k,l,m,n}}={\log }^{-1}5.137\\ {\text{N}}_{\text{i}\text{.j,k,l,m,n}}=0.137\times {10}^6\end{array}$

Calculate the second row of fourth column for allowable number of load applications are,

  $\begin{array}{l}\text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=}{\text{C}}_{\text{1}}{\left(\frac{{\text{S}}^{\text{i}}{}_{\text{c,i}}}{{\text{σ}}_{\text{i}\text{.j,k,l,m,n}}}\right)}^{{\text{C}}_{\text{2}}}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=2}{\left(\frac{\text{650}}{280}\right)}^{1.22}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=2}{\left(2.321\right)}^{1.22}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=2}\left(2.793\right)\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=}5.586\\ {\text{N}}_{\text{i}\text{.j,k,l,m,n}}={\log }^{-1}5.586\\ {\text{N}}_{\text{i}\text{.j,k,l,m,n}}=0.385\times {10}^6\end{array}$

Calculate the third row of fourth column for allowable number of load applications are,

  $\begin{array}{l}\text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=}{\text{C}}_{\text{1}}{\left(\frac{{\text{S}}^{\text{i}}{}_{\text{c,i}}}{{\text{σ}}_{\text{i}\text{.j,k,l,m,n}}}\right)}^{{\text{C}}_{\text{2}}}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=2}{\left(\frac{\text{650}}{275}\right)}^{1.22}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=2}{\left(2.363\right)}^{1.22}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=2}\left(2.855\right)\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=}5.710\\ {\text{N}}_{\text{i}\text{.j,k,l,m,n}}={\log }^{-1}5.710\\ {\text{N}}_{\text{i}\text{.j,k,l,m,n}}=0.512\times {10}^6\end{array}$

Calculate the fourth row of fourth column for allowable number of load applications are,

  $\begin{array}{l}\text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=}{\text{C}}_{\text{1}}{\left(\frac{{\text{S}}^{\text{i}}{}_{\text{c,i}}}{{\text{σ}}_{\text{i}\text{.j,k,l,m,n}}}\right)}^{{\text{C}}_{\text{2}}}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=2}{\left(\frac{\text{650}}{270}\right)}^{1.22}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=}5.840\\ {\text{N}}_{\text{i}\text{.j,k,l,m,n}}={\log }^{-1}5.840\\ {\text{N}}_{\text{i}\text{.j,k,l,m,n}}=0.691\times {10}^6\end{array}$

Calculate the fifth row of fourth column for allowable number of load applications are,

  $\begin{array}{l}\text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=}{\text{C}}_{\text{1}}{\left(\frac{{\text{S}}^{\text{i}}{}_{\text{c,i}}}{{\text{σ}}_{\text{i}\text{.j,k,l,m,n}}}\right)}^{{\text{C}}_{\text{2}}}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=2}{\left(\frac{\text{650}}{260}\right)}^{1.22}\\ \text{log}\left({\text{N}}_{\text{i}\text{.j,k,l,m,n}}\right)\text{=}5.840\\ {\text{N}}_{\text{i}\text{.j,k,l,m,n}}={\log }^{-1}6.116\\ {\text{N}}_{\text{i}\text{.j,k,l,m,n}}=1.308\times {10}^6\end{array}$

Allowable number of load applications are tabulated in the table.

• Determine total fatigue damage from $\text{Eq}\text{. 20}\text{.42}$of textbook "Traffic and Highway Engineering, 5th Edition" by using the formula,

  $\begin{array}{l}\text{D}{\text{I}}_{\text{F}}\text{=å}\frac{{\text{n}}_{\text{i,j,k,l,m,n}}}{{\text{N}}_{\text{i,j,k,l,m,n}}}\\ \text{i=age}\\ \text{j=month}\\ \text{k=axle type}\\ \text{l=load level}\\ \text{m=equivalent temperature difference between top and PCC surface}\\ \text{n=traffic offset path}\\ \text{o=hourle truck traffic}\end{array}$

  $\begin{array}{l}\text{For second row of column five is calculated as,}\\ \text{D}{\text{I}}_{\text{F}}\text{=}\frac{\text{0}\text{.02×1}{\text{0}}^{\text{6}}}{\text{0}\text{.137×1}{\text{0}}^{\text{6}}}\\ \text{D}{\text{I}}_{\text{F}}\text{=0}\text{.146}\end{array}$

For second row of column fiveis calculated as,

  $\begin{array}{l}\text{D}{\text{I}}_{\text{F}}\text{=}\sum \frac{{\text{n}}_{\text{i,j,k,l,m,n}}}{{\text{N}}_{\text{i,j,k,l,m,n}}}\\ \text{i=age}\\ \text{j=month}\\ \text{k=axle type}\\ \text{l=load level}\\ \text{m=equivalent temperature difference between top and PCC surface}\\ \text{n=traffic offset path}\\ \text{o=hourle truck traffic}\\ \text{D}{\text{I}}_{\text{F}}\text{=}\frac{0.05\times {10}^6}{0.385\times {10}^6}\\ \text{D}{\text{I}}_{\text{F}}\text{=}0.129\end{array}$

For third row of column fiveis calculated as,

  $\begin{array}{l}\text{D}{\text{I}}_{\text{F}}\text{=}\sum \frac{{\text{n}}_{\text{i,j,k,l,m,n}}}{{\text{N}}_{\text{i,j,k,l,m,n}}}\\ \text{D}{\text{I}}_{\text{F}}\text{=}\frac{0.08\times {10}^6}{0.512\times {10}^6}\\ \text{D}{\text{I}}_{\text{F}}\text{=}0.156\end{array}$

For fourth row of column fiveis calculated as,

  $\begin{array}{l}\text{D}{\text{I}}_{\text{F}}\text{=}\sum \frac{{\text{n}}_{\text{i,j,k,l,m,n}}}{{\text{N}}_{\text{i,j,k,l,m,n}}}\\ \text{D}{\text{I}}_{\text{F}}\text{=}\frac{0.10\times {10}^6}{0.691\times {10}^6}\\ \text{D}{\text{I}}_{\text{F}}\text{=}0.144\end{array}$

For fifth row of column fiveis calculated as,

  $\begin{array}{l}\text{D}{\text{I}}_{\text{F}}\text{=}\sum \frac{{\text{n}}_{\text{i,j,k,l,m,n}}}{{\text{N}}_{\text{i,j,k,l,m,n}}}\\ \text{D}{\text{I}}_{\text{F}}\text{=}\frac{0.12\times {10}^6}{1.308\times {10}^6}\\ \text{D}{\text{I}}_{\text{F}}\text{=}0.091\end{array}$

Therefore, total fatigue damage as shown in below table.

Load level	Number of applications(n)	Applied stresses,   ${\text{σ}}_{\text{i}\text{.j,k,l,m,n}}$   $\left(\text{lb/i}{\text{n}}^{\text{2}}\right)$	Allowable number of load applications	Total fatigue damage,   $\text{D}{\text{I}}_{\text{F}}$
$1$	$0.02\times {10}^6$	$300$	$0.137\times {10}^6$	$0.146$
$2$	$0.05\times {10}^6$	$280$	$0.385\times {10}^6$	$0.129$
$3$	$0.08\times {10}^6$	$275$	$0.512\times {10}^6$	$0.156$
$4$	$0.10\times {10}^6$	$270$	$0.691\times {10}^6$	$0.144$
$5$	$0.12\times {10}^6$	$260$	$1.308\times {10}^6$	$0.091$
				$\text{sum=0}\text{.666}$

• Determine the predicted amount of transverse cracking. Use $\text{Eq}\text{. 20}\text{.41}$ from textbook "Traffic and Highway Engineering, 5th Edition".

  $\begin{array}{l}\text{CRK=}\frac{1}{1+{\left(D{I}_F\right)}^{-1.98}}\\ \text{CRK=}\frac{1}{1+{\left(0.666\right)}^{-1.98}}\\ \text{CRK=}\frac{1}{1+2.236}\\ \text{CRK=}\frac{1}{3.236}\\ \text{CRK=}0.309\end{array}$

Conclusion:

Therefore, the expected speed predicted transverse cracking for the associated applied stresses and five traffic levels is $0.309$.