Question

M40 grade concrete is used for the construction of concrete pavement of thickness 28 cm. Poisson’s ratio of the concrete is 0.15. If the modulus of subgrade reaction is 8 kg/cm³, the radius of relative stiffness would be:

1. 93.186 cm
2. 52.15 cm
3. 98.15 cm
4. 111.186 cm
5.

Answer 1

Correct Answer - Option 1 : 93.186 cm

Concept:

Radius of Relative Stiffness:

In Westergard’s equations, the distance from the centre of the slab to the point of contra flexure is equal to the radius of relative stiffness for the interior loading condition.

It can be expressed as = \(l = \left \{ \dfrac{Eh^3}{12k(1-μ^2)} \right \}^{\frac{1}{4}}\)

E = Modulus of elasticity of cement concrete

h = Slab thickness

k = modulus of subgrade reaction

μ = Poisson's ratio

Calculation:

Given:

Grade of concrete = M40

∴ Modulus of elasticity of concrete = \(5000\sqrt f_{ck}\) 

= 5000 × \(\sqrt{40}\) = 31622.8 MPa = 31622.8 N/mm2 

= 316228 kg/cm3 

h = 28 cm, k = 8 kg/cm³ , μ = 0.15

So, \(l = \left \{ \dfrac{316228\times 28^3}{12\times 8(1-0.15^2)} \right \}^{\frac{1}{4}}\)

l = 92.7 cm

Option 1 is the nearest, hence answer = 93.186 cm