Pavement Design Multiple Choice Questions on “Highway Pavements – Design Factors – 2”.

1. The contact pressure of the wheel is obtained as ______

a) Force by wheel per pavement area

b) Load on wheel per contact area

c) Load on wheel per pavement area

d) Force by wheel per contact area

Answer: b

Clarification: The contact pressure of a wheel is defined as the ratio between the load on the wheel and the contact area. The contact area of the wheel with the pavement surface is considered.

2. If there are two wheels on one side of the axle, then it must be converted into ______

a) EWL

b) ESL

c) ESWL

d) EL

Answer: c

Clarification: ESWL stands for equivalent single wheel load. It is used to represent the load when more than one wheel is present in one side of the axle. The equivalent of both wheel loads is represented as ESWL.

3. In which of the traffic approaches for pavement design, is the thickness of pavement based on the number of repetitions of standard axle load?

a) Fixed traffic

b) Variable traffic

c) Variable vehicle

d) Fixed vehicle

Answer: d

Clarification: There are 3 methods for pavement design using traffic characteristics – fixed traffic, fixed vehicle and variable traffic and vehicle. In the fixed traffic method, the thickness of pavement is governed by a single load. In the fixed vehicle method, the thickness is governed by the number of repetitions of standard axle load. In the variable traffic and vehicle method, the loads are divided into groups and used for design.

4. What adverse effects does the term frost action relate to?

a) Frost heave and frost melting

b) Freeze/thaw cycles and frost melting

c) Frost heave and freeze/thaw cycles

d) Frost heave, freeze/thaw cycles and frost melting

Answer: d

Clarification: Frost action is a term that relates the adverse effects on the pavement. The adverse effects due to frost heave, frost melting and freeze/thaw cycles are included in the term. It is a sum total of all adverse effects due to the freezing temperatures on the pavement surface.

5. What is the formula for the centre to centre spacing between wheels, S?

a) S=(d-2a)

b) S=(d+a)

c) S=(d-a)

d) S=(d+2a)

Answer: d

Clarification: S=(d+2a) gives the right formula to determine the centre to centre spacing between the wheels of the vehicle. In the equation, d represents the clear gap between the wheels and a is the radius of the contact area of each wheel.

6. According to what criteria is the graphical method of ESWL computation carried out?

a) Deflection

b) Strain

c) Stress

d) Load

Answer: c

Clarification: The computation of ESWL can be carried out using two criteria. The first one is the deflection criteria and the second one is the stress criteria. For the graphical computation of ESWL, the stress criterion is used.

7. Which of the below is not used as salt as the remedy for frost action?

a) Zinc chloride

b) Sodium chloride

c) Sulphur dioxide

d) Calcium chloride

Answer: c

Clarification: Salts like zinc chloride, sodium chloride and calcium chloride are used to stabilize the soil and make them less prone to damage due to frost action. These salts work by reducing the freezing temperature of the water. Sulphur dioxide has antimicrobial property and is used as a preservative.

8. Which of the below is an incorrect assumption made in the Boyd and Foster method?

a) Equivalency concept based on equal deflection

b) Contact area is circular

c) Soil is elastic, homogeneous and isotropic

d) Influence angle is 45°

Answer: a

Clarification: In the Boyd and Foster method, the equivalency concept is based on equal stress and not on equal deflection. All other options represent the correct assumptions made by Boyd and Foster.

9. Thawing always proceeds from the bottom of the pavement to the top.

a) True

b) False

Answer: b

Clarification: The thawing can proceed from top to bottom or from bottom to top of the pavement or both ways. The whole process is dependent on the temperature of the pavement surface. The thawing may occur suddenly due to melting and it generally moves from the pavement surface downwards.

10. What would be the ESWL of a dual wheel load assembly of 4088 kg at a thickness of 30 cm, if the parameters S = 27 cm and d = 12 cm?

a) 3400.08 kg

b) 3040.08 kg

c) 3440.08 kg

d) 3404.08 kg

Answer: d

(frac{log,2P-log,P}{log,2s -log,{d/2}} = frac{log,P^{‘}-log,P}{log,30-log,{d/2}})

(frac{log,2×2044-log,2044}{log,2×27-log,frac{12}{2}}=frac{log,P^{‘}-log,2044}{log,30-log,frac{12}{2}})

(0.316=frac{log,P^{‘}-log,2044}{0.699})

(log,P^{‘}-log,2044=0.221)

(log,P^{‘}=0.221+log,2044=3.532)

(P^{‘}=antilog(3.532)=3404.08 kg)

11. As per the deflection criterion, the ESWL is that single wheel load having the same contact pressure which produces the same ______ at depth z.

a) Minimum deflection

b) Maximum deflection

c) Deflection

d) Threshold deflection

Answer: b

Clarification: The deflection criterion to obtain the ESWL states it as the single wheel load having the same contact pressure at which the value of the maximum deflection is the same at a depth z. The value of maximum deflection is taken into account here.

12. During cold weather, the bituminous layer of the pavement becomes ______ and cracks are formed.

a) Dry

b) Brittle

c) Soft

d) Moist

Answer: b

Clarification: The bituminous layers of the pavement become brittle during the cold weather. The bitumen is completely deprived of its flexible nature and thus it becomes brittle and this causes the cracks to develop as heavy wheel loads tread on it.

13. The elastic modulii values are determined using elasticity tests.

a) True

b) False

Answer: b

Clarification: It is necessary to ascertain the strength of various materials being used in the pavement, as they affect the design of pavement. The elastic modulii is a factor that gives a measure of the strength of materials. It is usually found out using the plate bearing test.

14. The temperature below freezing point and the duration of freezing are the factors that determine ______

a) Width of frost action

b) Speed of frost action

c) Depth of frost action

d) Intensity of frost action

Answer: c

Clarification: The depth of frost action is determined by two factors. The factors are the temperature below freezing point and the duration of freezing. The depth of frosts action is the parameter that gives the measure of the intensity of damage that can be caused by the same.

15. The strength parameter generally used for soil is ______

a) Cohesion

b) Proctor

c) Triaxial compression

d) CBR

Answer: d

Clarification: It is important to analyse the strength parameter of the various pavement materials before the design and construction of the pavement. The strength parameter of the subgrade soil is generally found out using the CBR test. The CBR value obtained is used to ascertain the strength characteristics of the subgrade soil.