Pavement Design written test Questions & Answers on “Highway Pavements – Design Factors – 3″.

1. If the frost heave is uniform, it doesn’t affect the pavement.

a) True

b) False

Answer: a

Clarification: The frost heave results in the lifting or swelling of the pavement surface. If the surface of the pavement has a differential rise due to frost heave, then the pavement will be damaged very badly. The differential rise happens because the centre of the pavement may rise higher than the sides or vice versa.

2. What would be the design repetitions for a period of 20 years equivalent to 2268 kg wheel load for the details of traffic survey conducted on a four-lane highway? The wheel load at which the design repetitions are to be found out is 3629 kg and the percentage of total traffic volume is 15.31%. The average daily traffic is found to be 250.

a) 1117630

b) 2235260

c) 6124

d) 279407

Answer: b

Clarification: The design repetition can be found out by multiplying the traffic for the design period of 20 years and the equivalent load factor. The equivalent load factor for 3629 kg is 8. The traffic for the design period is found out as:

Average daily traffic × Percentage of traffic × 365 × design period

So, for the load 3629 kg, the design repetitions would be 250×(frac{15.31}{100})×365×20×8=2235260

3. Which of the below is not an environmental factor to be considered for the design of pavements?

a) Depth of water table

b) Height of embankment

c) Depth of cutting

d) Formation width

Answer: d

Clarification: Formation width is the total width of the pavement, it is not an environmental factor. The environmental factors that affect the design of pavements can be related to the cutting and filling activities and the water table of the area. The height of embankment/filling, depth of cutting/excavation of the soil and the depth of water tale are considered.

4. What would be the ESWL of a dual wheel load assembly of 4088 kg at a thickness of 25 cm, if S = 25 cm and d = 8 cm?

a) 3372.87 kg

b) 3337.28 kg

c) 3708.75 kg

d) 3370.87 kg

Answer: a

Clarification: The ESWL can be found using the equation obtained from the interpretation of graphical analysis as below.

(frac{log ,2P-log ,P}{log ,2s-log ,{d/2}}=frac{log ,P^{‘}-log ,P}{log ,25-log ,{d/2}})

(frac{log ,2×2044-log ,2044}{log ,2×25-log ,frac{8}{2}}=frac{log ,P^{‘}-log ,2044}{log ,25-log ,frac{8}{2}})

(0.274=frac{log ,P^{‘}-log ,2044}{0.796})

(log ,P^{‘}-log ,2044=0.218)

(log ,P^{‘}=0.218+log ,2044=3.528)

(P^{‘}=antilog(3.528)=3372.87 kg)

5. The capillary cut off to arrest frost action can be done by ______

a) Geofibre

b) Laying pipes

c) Bituminous layer

d) Lowering the water table

Answer: c

Clarification: There are two ways to perform the capillary cut off to reduce the frost action. It can be done by inserting a layer of granular or bituminous material so that the rise of water can be stopped. Lowering the water table is not always possible. Laying pipes and use of geofibres are not employed as remedies for frost action.

6. What would be the centre to centre distance between the wheels if the clear gap between them is 70 mm and the radius of the contact area is 150 mm?

a) 370 cm

b) 370 mm

c) 230 cm

d) 230 mm

Answer: b

Clarification: The equation relating the clear gap (d), centre to centre spacing (S) and radius of contact area (r) is given by S=(d+2a). So, S can be found out as S=70+2×150=370 mm.

7. The formation of ice crystals on the pavement surface leads to the development of frost heave.

a) True

b) False

Answer: b

Clarification: The formation of ice crystals does lead to the frost heave, but it is not formed on the pavement surface. The ice crystals are formed due to the freezing of water that is held in the subgrade soil.

8. Which of the below cannot be computed using the plate bearing test?

a) Subgrade modulus

b) Elastic moduli of base course

c) Resilient modulus of soil

d) Elastic moduli of sub-base course

Answer: c

Clarification: The strength evaluation of pavement materials is based on two things – CBR and elastic modulii. To find the elastic modulii, plate bearing test and triaxial compression tests are used. Triaxial compression test gives the resilient modulus of the subgrade soil.

9. The wheel load and the corresponding number of repetitions to failure are 2268 kg – 105000 and 4082 kg – 6500 respectively. What is the equivalent load factor for the load 4082 kg?

a) 8

b) 64

c) 16

d) 32

Answer: c

Clarification: Equivalent wheel load is the equivalency of the load under consideration to the 2268 kg load. The number of repetitions of load 2268 kg i.e. 105000 is divided by the number of repetitions corresponding to the load under consideration i.e. 6500. The equivalent wheel load is hence obtained as 105000/6500 = 16.15. It is written as the closest power of 2, so the equivalent load factor would be 16.

10. What would be the design repetitions for a period of 20 years equivalent to 2268 kg wheel load for the details of traffic survey conducted on a two-lane highway? The wheel load at which the design repetitions are to be found out is 4082 kg and the percentage of total traffic volume is 7.66%. The average daily traffic is found to be 210.

a) 1878784.8

b) 1878844.8

c) 1877884.8

d) 1887744.8

Answer: b

Clarification: The design repetition is the product of the traffic for the design period of 20 years and the equivalent load factor. The equivalent load factor for 4082 kg is 16. The traffic for the design period is found out as:

Average daily traffic × Percentage of traffic × 365 × design period

Therefore, the design repetitions for the load 4082 kg would be 210×(frac{7.66}{100})×365×20×16=1878844.8

11. The variation in the moisture content of the subgrade is not dependent on which factor?

a) Temperature

b) Water table

c) Pavement type

d) Vapour movement

Answer: a

Clarification: Temperature is not a parameter affecting the moisture content of the subgrade, it is a factor affecting the design of pavements. The fluctuations in the water table, the type of pavement and shoulder, the capillary action and vapour movement influence the moisture content in the subgrade.

12. The stress on the pavement surface due to bullock carts are ______

a) Negligible

b) High

c) Little

d) Moderate

Answer: b

Clarification: The wheels of a modern bullock cart are steel wheels. When these wheels tread the pavement, they impose high stress on the pavement surface. While the stress is negligible in the bottom layers due to their small gross weight.

13. What is the wheel configuration of the trailer portion of a tractor-trailer unit?

a) Single axle

b) Dual axle

c) Triple axle

d) Tandem axle

Answer: d

Clarification: The trailer portion carries the major load portion of the tractor-trailer unit. The axle provided is the tandem axle which consists of two axles on which two wheels are placed on one side of each axle. The tractor part consists of a single axle single wheel and single axle dual wheel assembly.

14. Soils having ______ result in variation in volume with variation in moisture.

a) High plasticity

b) High liquidity

c) Low plasticity

d) Low liquidity

Answer: a

Clarification: Liquidity of a soil indicates the natural moisture content present in the soil. Soils having high plasticity are prone to a greater extent of swelling and shrinking than those with low plasticity and thus they tend to damage the pavement.

15. What is the reliability value for flexible pavements of state highways in India?

a) 95%

b) 95

c) 75%

d) 75

Answer: a

Clarification: As per AASHTO, the reliability value gives for flexible pavement in India is as follows:

National highways – >>95%

State highways and major district roads – 85 to 95%

Village roads and other district roads – 50 to 75%

Reliability value is expressed as a percentage and it gives the probability that particular damage to the pavement would remain under permissible limits during its design period.