Question

In a plane cement concrete pavement of thickness 20 cm, if the coefficient of friction is 1.4 and unit weight of cement concrete is 2400 kg/m${}^3$, allowable tensile stress is 1 kg/cm${}^2$ the spacing between the contraction joints is ___________m.

Answer 1

$\text{Spacing between the contraction joints}=\frac{2S_c}{Wf}$
$=\frac{2\times 1\times {10}^4}{(2400\times 1.4)}=5.95m$
As per IRC maximum spacing is 4.5 m. Hence adopt spacing of 4.5 m .