Also determine the probable location where the crack is likely to develop due to corner loading.

Wheel load, P = 5100 kg

Modulus of elasticity of cement concrete, E = 3.0 x $10^5 kg/cm^2$

Pavement thickness, h = 18 cm

Poisson's ratio of concrete, $\mu$ = 0.15

Modulus of subgrade reaction, K = 6.0 $kg/cm^3$

Radius of contact area, a = 15 cm

Radius of relative stiffness(l) is given by

$l = [\frac{Eh^3}{12K(1 - \mu^2)}]^\frac{1}{4} = 70.6\hspace{0.05cm}cm$

The equivalent of resisting section is given by

a/h = 15/18 = 0.833 < 1.74

b = $\sqrt{1.6a^2 + h^2} - 0.675h$ = 14.0 cm

(a) Stress at the interior:

$S_i = \frac{0.316 P}{h^2}[4log_{10}(\frac{l}{b}) +1.069] = 19.3 kg/cm^2$

(b) Stress at the edge:

$S_e = \frac{0.572P}{h^2}[4log_{10}(\frac{l}{b}) + 0.359] = 28.54 kg/cm^2$

(c) Stress at the corner:

$S_c = \frac{3P}{h^2}[1-(\frac{a\sqrt{2}}{l})^{0.6}] = 24.27 kg/cm^2$

(d) Location where corner load crack develops

Location where the crack is likely to develop due to corner loading, the distance from the corner of the slab

$X = 2.58\sqrt{al} = 2.58\sqrt{(15\times70.6)} = 83.96 = 84\hspace{0.05cm}cm$