Pavement Design Questions and Answers – Highway Pavements – Design Factors – 1

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This set of Pavement Design Multiple Choice Questions & Answers (MCQs) focuses on “Highway Pavements – Design Factors – 1”.

1. Which of the below design factor and its effect are matched incorrectly?
a) High stress – high thickness
b) High stability – less thickness
c) Environmental factor – frost action
d) Soil property – stability of subgrade
View Answer

Answer: c
Explanation: The problem due to frost is a climatic factor that affects the design of pavements. The thickness of pavement can be reduced when the stability is high and stress is low. The stability of subgrade is very much dependent on the type of soil that is used.
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2. What is the factor that the design of semi-rigid pavements is based on?
a) Stability
b) Cracks
c) Stress
d) Deflections
View Answer

Answer: b
Explanation: The design of semi-rigid pavements can be carried out only after assessing the pavement conditions under shrinkage cracks. It is necessary to know the pattern, mode of propagation and fatigue behaviour under hair cracks.

3. Wheel load configuration helps in knowing the ______ required for the design of pavements.
a) Number of wheels
b) Load distribution
c) Contact pressure
d) Load factor
View Answer

Answer: b
Explanation: The wheel load configuration gives the pattern in which the wheel load would be applied on the pavement surface. It gives the number of wheels, but that is not enough for the design process.

4. The ratio of contact pressure to tyre pressure is called as ______
a) Pressure ratio
b) Load factor
c) Rigidity factor
d) Power ratio
View Answer

Answer: c
Explanation: Rigidity factor gives the ratio between contact pressure and tyre pressure. The value of the rigidity factor depends on the degree of tension that is developed between the walls of the tyres. Pressure ratio is a term used in aeronautical engineering, the power ratio is related to business and load factor is related to aircraft.

5. Fast-moving vehicles tend to impose more damage on the pavement.
a) True
b) False
View Answer

Answer: b
Explanation: The vehicles moving at creep speed tend to impose a higher amount of damage to the pavement than the vehicles moving at high speed. When the vehicle moves slowly, the load on the pavement lasts for a longer time and leads to deformations.
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6. What is the maximum axle load allowed on highways in India?
a) 6780 kg
b) 4085 kg
c) 2400 kg
d) 8170 kg
View Answer

Answer: d
Explanation: The maximum legal axle load allowed on highways in India is 8170 kg as per IRC guidelines. The maximum ESWL allowed is 4085 kg.

7. Temperature affects both the flexible and rigid pavement in different ways.
a) True
b) False
View Answer

Answer: a
Explanation: Temperature changes can affect the flexible pavement in a way different from that of rigid pavement. The bituminous layers in the flexible pavement are subjected to being brittle or easily deformable. The rigid pavement slab is subjected to warping stresses.

8. Calculate the ESWL at a thickness of 200 mm for a dual wheel load assembly carrying 2044 kg. The centre to centre spacing between tyre is 250 mm and the clear gap between tyres is 100 mm.
a) 3100.65 kg
b) 3000.65 kg
c) 3100.85 kg
d) 3000.85 kg
View Answer

Answer: c
Explanation: The ESWL and thickness can be plotted on a semi-log graph and the results can be interpreted from that. The thickness corresponding to ESWL 2P and P are 2S and d/2 respectively. So, for a thickness of 200 mm, the ESWL would be assumed as P’. Considering the slope of the linear graph obtained, the relation between these terms can be written as follows.
\(\frac{log \,2P-log \,P}{log \,2s-log \,{d/2}}=\frac{log \,P^{‘}-log \,P}{log \,200-log \,{d/2}}\)
Therefore, the values can be substituted in the equation as
\(\frac{log \,2×2044-log \,2044}{log \,2×250-log \,\frac{100}{2}}=\frac{log \,P^{‘}-log \,2044}{log \,200-log \,\frac{100}{2}}\)
\(0.301=\frac{log \,P^{‘}-log \,2044}{0.602}\)
\(log \,P^{‘}-log \,2044=0.181\)
\(log \,P^{‘}=0.181+log \,2044=3.49\)
\(P^{‘}=antilog(3.49)=3100.85 kg\)

9. Tyre pressure is exactly the same as ______ pressure.
a) Contact
b) Force
c) Wheel
d) Inflation
View Answer

Answer: d
Explanation: Tyre pressure, contact pressure and inflation pressure are related to each other. Tyre pressure and inflation pressure essentially are the same thing. Contact pressure and tyre pressure vary according to the standard value of pressure.
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10. Who proposed the widely accepted rule of thumb criterion for identifying frost susceptible soils?
a) Casagrande
b) Berggren
c) Stefan
d) Burt
View Answer

Answer: a
Explanation: Dr Arthur Casagrande is the person who proposed the rule of thumb criterion for identifying a soil that is susceptible to frost. Berggren and Stefan formulas are used to determine the temperature changes in the soil mass.

11. Compute the clear gap between the wheels if the centre to centre spacing is 280 mm and the radius of contact area is 110 mm.
a) 50 cm
b) 50 mm
c) 60 mm
d) 60 cm
View Answer

Answer: c
Explanation: 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, d can be found out as d=S-2a=280-2×110=60 mm.

12. What is the original shape of the contact area?
a) Circle
b) Ellipse
c) Conical
d) Triangular
View Answer

Answer: b
Explanation: The original shape of the contact area is elliptical. It is considered to be circular for the ease of carrying out the calculations. It is used in the calculation of contact pressure.

13. Which of the below is not a factor that the frost action depends upon?
a) Water table
b) Temperature
c) Soil type
d) Vegetation cover
View Answer

Answer: d
Explanation: The factor that affects the frost action is the cover or topping that is provided on top of the pavement. It refers to the white or black topping. The other factors that influence the frost action are the frost susceptible soil, depressed temperature below freezing point and the supply of water.
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14. The wheel load and the corresponding number of repetitions to failure are 2268 kg – 105000 and 3175 kg – 22500 respectively. What is the equivalent load factor for the load 3175 kg?
a) 6
b) 2
c) 8
d) 4
View Answer

Answer: d
Explanation: The equivalent wheel load is obtained by considering the equivalency of the load to the 2268 kg load. It is obtained by dividing 105000 by the number of repetitions corresponding to the load that is considered. The equivalency in the question is 105000/22500 = 4.66. The equivalent wheel load can be written as the closest power of 2, i.e. the equivalent load factor would be 4.

15. What is the rigidity factor if the average tyre pressure is 7 kg/cm2?
a) 0
b) 1
c) -1
d) 2
View Answer

Answer: b
Explanation: Rigidity factor is the ratio between contact pressure and tyre pressure. When the tyre pressure is 7 kg/cm2, it is equal to the contact pressure. When the tyre pressure is less than the contact pressure, the rigidity factor is found to be higher than one. When the tyre pressure is more than the contact pressure, the rigidity factor is less than one.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn