Question

A CC pavement of 4 m width and of thickness 25 cm is reinforced with 10 mm dia bars at 25 cm. If the friction factor f = 1.4 and allowable tensile stress 1400 kg/sq.cm, spacing between the contraction joints is (unit weight of concrete is 2400 kg/cum)

• A. 13.56 m
• B. 9.16 m
• C. 10.467 m
• D. 5.2 m

Answer 1

The correct option is C 10.467 m

Area of steel in one direction of slab,
$=\frac{4}{0.25}\times \frac{\pi \times 1^2}{4}=12.56c{m}^2$

Spacing between the contraction joints is obtained by equating the force,
$\gamma \times b\times \frac{L_c}{2}\times h\times f={\sigma }_s{A}_s$
$2400\times 4\times \frac{L_c}{2}\times 0.25\times 1.4=1400\times 12.56$
$L_c=10.467m$