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