Determine the spacing between contraction joints for a 3.5 metre slab width having a thickness of 20 cm and coefficient of friction of 1.5 for plain cement concrete (unit weight = 2400 kg/cm²) and allowable stress in tension in cement concrete = 0.8 kg/cm².

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BHEL Engineer Trainee Civil 24 Aug 2023 Official Paper

Answer 1

Concept:

Equating frictional force with the tensile force resisted by the pavement:

Frictional force, \( F = f \times \frac{\gamma \cdot B \cdot h}{2} \)

Tensile force resisted = \( B \cdot h \cdot \sigma_c \)

Given:

Width of slab, B = 3.5 m

Thickness, h = 0.2 m

Unit weight of concrete, \( \gamma = 24 \, \text{kN/m}^3 \)

Coefficient of friction, f = 1.5

Allowable tensile stress in concrete, \( \sigma_c = 80 \, \text{kN/m}^2 \)

Calculation:

Equating both forces:

\( 1.5 \times \frac{24 \cdot l_c \cdot 3.5 \cdot 0.2}{2} = 3.5 \cdot 0.2 \cdot 80 \)

Solving:

\( 1.5 \times \frac{24 \cdot l_c \cdot 0.7}{2} = 56 \)

\( 1.5 \times 8.4 \cdot l_c = 56 \Rightarrow l_c = \frac{56}{12.6} = 4.44 \, \text{m} \)