Question

Consider the following data with respect to the design of flexible pavement:

Design wheel load = 4200 kg

Tyre pressure = 6.0 kg/cm²

Elastic modulus = 150 kg/cm²

Permissible deflection = 0.25 cm

(take π^1/2 = 1.77, π^-1/2 = 0.564, \(\frac{1}{\pi } = 0.318,\) and π² = 9.87)

The total thickness of flexible pavement for a single layer elastic theory will be nearly
1. 42 cm
2. 47 cm
3. 51 cm
4. 56 cm

Answer 1

Correct Answer - Option 3 : 51 cm

Concept:

Thickness of pavement by single layer elastic theory is given by,

\(T = \sqrt {{{\left( {\frac{{3P}}{{2\pi {E_s}{\rm{\Delta }}}}} \right)}^2} - {a^2}}\) 

Contact pressure, \(\left( p \right) = \frac{P}{{\pi {a^2}}}\)

Where, P = Design wheel load (in kg)

E_s = Elastic modulus (in kg/km²)

a = radius of contact area (in cm)

Δ = permissible deflection (in cm)

Calculation:

Design wheel load = 4200 kg

Tyre pressure = 6.0 kg/cm²

Elastic modulus = 150 kg/cm²

Permissible deflection = 0.25 cm

\(p = \frac{P}{{\pi {a^2}}} \Rightarrow 6 = \frac{{4200}}{{\pi \times {a^2}}} \Rightarrow a = {\left( {\frac{{4200}}{{6 \times \pi }}} \right)^{1/2}} = 14.92\;cm\) 

\(T = \sqrt {{{\left( {\frac{{3 \times 4200}}{{2\pi \times 150 \times 0.25}}} \right)}^2} - {{\left( {14.92} \right)}^2}} = 51.352\)

= 51.352 cm

= 51 cm