Pavement Design Questions and Answers – McLeod and Burmister Method – 2

This set of Pavement Design Multiple Choice Questions & Answers (MCQs) focuses on “McLeod and Burmister Method – 2”.

1. Burmister’s method of pavement design is a type of ______ method.
a) Empirical
b) Analytical
c) Semi-empirical
d) Theoretical
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2. In which year did McLeod present his method of flexible pavement design?
a) 1979
b) 1983
c) 1969
d) 1981
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3. What is the equation for deflection under flexible pavement according to Burmister’s method?
a) \(Δ=\frac{3paF_2}{E_p}\)
b) \(Δ=\frac{1.5paF_2}{E_p}\)
c) \(Δ=\frac{3paF_2}{E_s}\)
d) \(Δ=\frac{1.5paF_2}{E_s}\)
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4. The subgrade support in the McLeod method is calculated for ______ diameter plate and ______ deflection.
a) 60 cm, 0.25 cm
b) 60 cm, 0.5 cm
c) 30 cm, 0.25 cm
d) 30 cm, 0.5 cm
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5. The deflection factor in the equation for Burmister’s method depends on that ratio of ______
a) Modulus of rigidity
b) Wheel load
c) Modulus of elasticity
d) Particle size
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6. Find the elastic modulus of soil when it is subjected to a plate bearing test. The diameter of plate used is 30 cm and the pressure yielded at 0.5 cm deflection is 1.25 kg/cm².
a) 44.25 kg/cm²
b) 45 kg/cm²
c) 44 kg/cm²
d) 45.25 kg/cm²
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7. Design a pavement using Burmister’s approach for a wheel load of 4100 kg and tyre pressure of 6 kg/cm² for an allowable deflection of 0.5 cm. The plate load test was conducted using a 30 cm diameter plate and the following observations were obtained:
For subgrade, pressure at 0.5 cm deflection = 1.2 kg/cm²
Over 15 cm thick base course, pressure at 0.5 cm deflection = 3.8 kg/cm²
a) 36.33 cm
b) 34.33 cm
c) 34.66 cm
d) 33.66 cm
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Answer: c
Explanation: The first step is to find the elastic modulus of the subgrade using rigid plate equation and considering single layer and hence the deflection factor, F = 1.
\(Δ=\frac{1.18paF_2}{E_s}\)
\(E_s=\frac{1.18paF_2}{Δ}=\frac{1.18×1.2×15×1}{0.5}=42.48 kg/cm^2\)
In order to obtain the elastic modulus ratio, we need to calculate the deflection factor assuming that two-layer in the equation for a rigid plate.
\(Δ=\frac{1.18paF_2}{E_s}\)
\(F_2=\frac{ΔE_s}{1.18pa}=\frac{0.5×42.48}{1.18×3.8×15}\)=0.316
The design chart developed by Donald M. Burmister can be used to find the necessary parameters. A sample of the same chart with values and data pertaining to the question is shown below. All the computations have been represented in one chart.
For F₂ = 0.316 and z/a = 15/15 = 1, the ratio of elastic modulus of subgrade and pavement can be obtained as 50. This is shown using the marking A in the design chart.

The next step is to find the deflection under a flexible plate. The equation for that is
\(Δ=\frac{1.15paF_2}{E_s}\)
In the equation,
\(a=(\frac{P}{pπ})^{\frac{1}{2}}=(\frac{4100}{6×π})^{1/2}\)=14.75 cm
We have to find the deflection factor,
\(F_2=\frac{ΔE_s}{1.5pa}=\frac{0.5×42.48}{1.5×6×14.75}\)=0.16
Using this F₂=0.16 and ratio 50, the z/a ratio can be found from the design chart as in the figure above. Marking B is used to show the same.
For the above values, the z/a ratio from the graph is obtained as 2.35. Therefore, the thickness of the pavement z = 2.35 X 14.75 = 34.66 cm.

8. In the equation for deflection for rigid and flexible plate using Burmister’s method, the Poisson’s ratio for both soil and pavement material is the same.
a) True
b) False
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9. Burmister developed his method of design in collaboration with the Canadian Department of Transport.
a) True
b) False
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10. Design a pavement using Burmister’s approach for a wheel load of 4100 kg and tyre pressure of 6.5 kg/cm² at an allowable deflection of 0.5 cm. The plate load test was conducted using a 30 cm diameter plate and the pressure at 0.5 cm deflection for the subgrade was found out to be 1.4 kg/cm² and the pressure over 20 cm thick base course 0.5 cm deflection is obtained as 4 kg/cm².
a) 45.05 cm
b) 54.05 cm
c) 56.05 cm
d) 46.05 cm
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Answer: d
Explanation: The first step is to find the elastic modulus of the subgrade using rigid plate equation. Single layer is considered and hence the deflection factor, F = 1.
\(Δ=\frac{1.18paF_2}{E_s}\)
\(E_s=\frac{1.18paF_2}{Δ}=\frac{1.18×1.4×15×1}{0.5}=49.56 kg/cm^2\)
In order to obtain the elastic modulus ratio, we need to calculate the deflection factor in the equation for the rigid plate by assuming that the pavement is a two-layer system.
\(Δ=\frac{1.18paF_2}{E_s}\)
\(F_2=\frac{ΔE_s}{1.18pa}=\frac{0.5×49.56}{1.18×4×15}\)=0.35
The design chart developed by Donald M. Burmister is used to find the necessary parameters. A sample of the chart with data pertaining to the question is as shown below. All the computations have been represented in the same chart.
Using the design chart, for F₂ = 0.35 and z/a = 20/15 = 1.33, the ratio of elastic modulus of subgrade and pavement can be obtained as 16. This is shown in the design chart using marking A.

The next step is to find the deflection under a flexible plate. The equation for that is
\(Δ=\frac{1.5paF_2}{E_s}\)
In the equation,
\(a=(\frac{P}{pπ})^{\frac{1}{2}}=(\frac{4100}{6.5×π})^{1/2}\)=14.17 cm
We have to find the deflection factor,
\(F_2=\frac{ΔE_s}{1.5pa}=\frac{0.5×49.56}{1.5×6.5×14.17}\)=0.18
Using this F₂=0.18 and ratio 16, the z/a ratio can be found from the design chart as in the above figure. Marking B can be used to understand the same.
From the chart, the z/a ratio is obtained as 3. Therefore, the thickness of the pavement z = 3.25 X 14.17 = 46.05 cm.

Sanfoundry Global Education & Learning Series – Pavement Design.

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