Design wheel load = 7500kg, Contact pressure = $7.5kg/cm^2$, Spacing between longitudinal joints = 3.75m and contractions joints is 4.2m, Elastic modulus of the pavement materials/CC is $3\times10^5kg/cm^2$, Poissons ratio = 0.15, Modulus of sub gradereaction = $30kg/cm^3$. Thermal coefficient of cc per $^\circ C=1\times10^{-5}/CC$, flexural strength of CC = $45kg/cm^2$, Max. temperature differential at the location for pavement thickness values of 24.26 and 30 cm are respectively 15.6, 16.2 and 16.8 $^\circ C$. Calculate the desired factor of safety with respect to load stress and warping stress at edge region.

(1)

$\textit{To find radius (l) of relative stiffness:}$

$P = \textit{Design wheel load} = 7500\hspace{0.05cm}kg$

$p = \textit{Contact pressure} = 7.5\hspace{0.05cm}kg/cm^2$

$\textit{We know},$

$a = \textit{Radius of load} = \sqrt{\frac{P}{p \pi}}$

$\hspace{03cm} = \sqrt{\frac{7000}{7.5\pi}} = 17.24\hspace{0.05cm}cm$

$\textit{TRIAL 1 : Assume pavement thickness (h) as 25 cm}$

$\hspace{02cm}\therefore h = 25\hspace{0.05cm}cm$

$\textit{Given: subgrade reaction}, K = 30\hspace{0.05cm}k/cm^3 , E = 3\times10^5\hspace{0.05cm}kg/cm^2, \mu = \textit{Poisson's ratio} = 0.15$

$\textit{We know},$

$\hspace{0.5cm}\textit{Radius of relative stiffness}, l = [\frac{Eh^3}{12K(1 - \mu^2)}]$

$\hspace{03cm} = [\frac{3\times10^5\times(15)^3}{12\times30(1- 0.15^2)}]$

$\hspace{2.5cm}\therefore l = 60.41\hspace{0.05cm}cm$

2.

$\hspace{0.05cm} \textit{To find radius of resisting section} (b)$

$\hspace{0.5cm}\frac{a}{h} = \frac{17.24}{25} = 0.69 \lt 1.724$

Hence substituting the values of a and h in the equation of radius of resisting section, we get

$\hspace{0.5cm}\textit{Radius of resisting section}, b = \sqrt{1.6a^2 + h^2} - 0.67h$

$\hspace{03cm} = \sqrt{1.6\times17.24^2 + 25^2} - 0.67\times25$

$\hspace{2.5cm}\therefore b = 16.29\hspace{0.05cm}cm$

$3.\hspace{0.05cm}\textit{To find} S_e$

$\hspace{0.5cm}(\frac{l}{b}) = \frac{60.41}{16.29} = 3.71$

$\textit{Substituting the value of (\frac{l}{b} in following equation of S_e}$

$\hspace{0.5cm} \therefore S_e = \frac{0.572 P}{h^2}[4 log_{10} \frac{l}{b} + 0.359]$

$\hspace{0.5cm} \therefore S_e = 16.98\hspace{0.05cm}kg/cm^2$

$4.\hspace{0.05cm}\textit{ To find flexural strength of rigid pavement}$

$\textit{For h = 25 cm, temperature differential by interpolation}$

$\hspace{02cm} = \frac{(15.6 + 16.2)}{2} = 15.9^\circ C$

$\hspace{01cm}L_x = 4.2\hspace{0.05cm}m = 420\hspace{0.05cm}cm, L_y = 375\hspace{0.05cm}cm$

$\textit{Warping stress for the higher ratio};$

$\frac{L_x}{l} = \frac{420}{60.41} = 6.96$

From Bradbury's warping stress coefficient chart as shown in fig. corresponding to

$\frac{L_x}{l} = 6.96 and C_s = 0.99$

$e = 1\times10^{-5} per ^\circ C$

$\textit{Warping stress at edge}$

$S_{t(e)} = \frac{C_x . E . e . t}{2}$

$\therefore S_{t(e)} = 23.61$

$\therefore \textit{Total flexural stress} = [S_e + S_{t(e)}] = 45\hspace{0.05cm}kg/cm^2$

$\therefore \textit{Flexural strength of rigid pavement} = 45\hspace{0.05cm}kg/cm^2$

$5. \textit{To find factor of safety}$

$\hspace{02cm}\textit{Factor of safety} = \frac{\textit{Flexural strength}}{\textit{Total flexural stress}}$

$\hspace{03cm} = \frac{45}{40.59} = 1.109$

Since factor of safety obtained is in between the required range of 1.1 and 1.2 adopt the design thickness of rigid pavement as 25 cm

$\therefore \textit{Design thickness of rigid pavement} = 25\hspace{0.05cm}cm$