This set of Pavement Design Assessment Questions and Answers 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

View Answer

Explanation: Burmister’s method used to design of flexible pavement comes under the semi-empirical method of design. The method uses the plate load test to get parameters and has an equation to get the values required to compute the pavement thickness.

2. In which year did McLeod present his method of flexible pavement design?

a) 1979

b) 1983

c) 1969

d) 1981

View Answer

Explanation: McLeod presented his ideas in the year 1969. It was adapted by the Asphalt Institute in the year 1979 and again modified in 1983. The method was also adapted by the Asphalt Emulsion Manufacturers Association in the year 1981.

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) \(Δ=\frac{1.5paF_2}{E_s}\)

View Answer

Explanation: \(Δ=\frac{1.5paF_2}{E_s}\) the equation used to find the deflection of flexible pavement using Burmister’s method. For a flexible pavement, the constant used is 1.5 and the modulus of elasticity of the subgrade is considered.

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

View Answer

Explanation: Subgrade support is used in the McLeod method for calculation of the thickness of the pavement. It is done by using the plate load test. For the test, the plate of 30 cm diameter and a deflection of 0.5 cm is considered. The number of repetitions for the test is taken as ten.

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

View Answer

Explanation: The equation used to calculate the thickness of pavement using the Burmister’s method makes use of a deflection factor F and it depends on the ratio of modulus of elasticity of subgrade soil to that of the pavement.

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/cm^{2}.

a) 44.25 kg/cm^{2}

b) 45 kg/cm^{2}

c) 44 kg/cm^{2}

d) 45.25 kg/cm^{2}

View Answer

Explanation: The equation for elastic modulus can be formulated from the equation for deflection in a rigid plate in Burmister’s method. The equation is \(Δ=\frac{1.18paF_2}{E_s}\). It is assumed as a single layer, so F = 1.

0.5=\(\frac{1.18×1.25×15×1}{E_s}\)

0.5=\(\frac{22.125}{E_s}\)

\(E_s=\frac{22.125}{0.5}=44.25\) kg/cm

^{2}

7. Design a pavement using Burmister’s approach for a wheel load of 4100 kg and tyre pressure of 6 kg/cm^{2} 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/cm^{2}

Over 15 cm thick base course, pressure at 0.5 cm deflection = 3.8 kg/cm^{2}

a) 36.33 cm

b) 34.33 cm

c) 34.66 cm

d) 33.66 cm

View Answer

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 F

_{2}= 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 F

_{2}=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

View Answer

Explanation: In order to derive the Burmister’s equation from Boussinesq’s equation, the Poisson’s ratio for the soil and the pavement material were taken to be the same. It was taken as 0.5 and this simplifies the equation and reduces it to the required form.

9. Burmister developed his method of design in collaboration with the Canadian Department of Transport.

a) True

b) False

View Answer

Explanation: It was McLeod who developed his method by using the plate load test conducted on airfield and highway pavements by the Canadian Department of Transport. Burmister, on the other hand, developed his method by considering the pavement as a layered system.

10. Design a pavement using Burmister’s approach for a wheel load of 4100 kg and tyre pressure of 6.5 kg/cm^{2} 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/cm^{2} and the pressure over 20 cm thick base course 0.5 cm deflection is obtained as 4 kg/cm^{2}.

a) 45.05 cm

b) 54.05 cm

c) 56.05 cm

d) 46.05 cm

View Answer

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 F

_{2}= 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 F

_{2}=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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