Pavement Design MCQ Quiz - Objective Question with Answer for Pavement Design - Download Free PDF
Last updated on Jun 11, 2025
Latest Pavement Design MCQ Objective Questions
Pavement Design Question 1:
The critical condition of stresses for combination of stresses in cement concrete pavement during summer is :
Answer (Detailed Solution Below)
Pavement Design Question 1 Detailed Solution
Explanation:
Combination of stresses:
(i) There is no frictional stress at the corner region. Out of various wheel stresses
- Corner stress is minimum as there is a discontinuity in both direction
- Interior stress is maximum
- Edge stress is in intermediate-range
(ii) Temperature stress is critical at the edge and interior and it is minimum at corner. At the corner, resistance due to weight is minimum, hence warping stress is minimum.
(iii) In combination of wheel load and temperature, edge region is mos critical, hence designing is done using edge region stress and however checking is done for corner region.
Critical cases of stress combination:
(i) Summer mid-day: The critical stress for edge region is given by
Scritical = Sedge + Stemperature - Sfrictional
(ii) Winter, mid-day: The critical combination of stress for the edge region is given by
Scritical = Sedge + Stemperature + Sfrictional
(iii) Mid-nights: The critical combination of stress for the corner region is given by
Scritical = Sedge + Stemperature
Pavement Design Question 2:
The interface treatment provided to plug in the voids of porous surfaces and to bond loose particles in bituminous pavements is called
Answer (Detailed Solution Below)
Pavement Design Question 2 Detailed Solution
Explanation:
Prime coat:
The prime coat is an application of low viscous cutback bitumen to an absorbent surface like granular bases on which binder layer is placed. It provides bonding between two layers. Unlike the tack coat, prime coat penetrates into the layer below, plugs the voids, and forms a watertight surface.
Seal Coat:
Seal Coat Seal coat is a thin surface treatment used to water-proof the surface and to provide skid resistance.
Tack coat:
Tack coat is a very light application of asphalt, usually asphalt emulsion diluted with water. It provides proper bonding between two layers of binder course and must be thin, uniformly cover the entire surface, and set very fast.
The bituminous prime coat is the first application of low viscosity liquid bituminous material over an existing porous or absorbent pavement surface like the WBM base course.
Pavement Design Question 3:
The highest CBR number is required for
Answer (Detailed Solution Below)
Pavement Design Question 3 Detailed Solution
Explanation:
The California Bearing Ratio (CBR) is a test used to determine the strength of subgrade and base materials for road and pavement design. A higher CBR value indicates better load-bearing capacity. Among the layers in a pavement system, the subgrade must possess the highest CBR because it serves as the foundation for all upper layers.
Pavement
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Refers to the entire road structure, including surface, base, sub base, and subgrade.
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It is not a single layer, hence not directly assigned a CBR value.
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CBR testing is conducted for individual layers like subgrade or sub base.
Sub grade
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This is the bottom-most layer of the pavement structure.
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It bears the entire load transmitted from the upper layers (sub base, base, and pavement surface).
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A weak subgrade leads to settlement and failure of the road structure.
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Requires the highest CBR to ensure long-term pavement performance.
Sub base
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Lies directly above the subgrade.
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Distributes the load from the base course to the subgrade.
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Requires moderate CBR, but not as high as the subgrade.
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Helps in drainage and frost protection.
Base
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Sits above the sub base and below the surface course.
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Made of high-quality material (e.g., crushed stones).
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Provides structural support and improves load distribution.
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Has high CBR requirements but still relies on a strong subgrade.
Pavement Design Question 4:
The maximum thickness of expansion joint in rigid pavements is
Answer (Detailed Solution Below)
Pavement Design Question 4 Detailed Solution
Explanation:
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The maximum thickness of expansion joints in rigid pavements is typically limited to 25 mm as per common design standards (like IRC guidelines).
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This thickness provides enough room for concrete expansion due to temperature changes while maintaining structural integrity.
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Expansion joints help prevent cracks and damage by allowing controlled movement between concrete slabs.
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Keeping the joint thickness within this limit ensures durability and reduces maintenance needs.
Additional Information
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Expansion joints are gaps placed between concrete slabs in rigid pavements to accommodate expansion and contraction caused by temperature changes.
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They prevent the development of stresses that can cause cracking or damage to the pavement structure.
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The joints are usually filled with compressible materials like bitumen, rubber, or foam to allow movement while preventing debris entry.
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Proper sizing and spacing of expansion joints are critical to maintain pavement durability and performance.
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Typically, the maximum thickness of these joints is kept around 25 mm to balance flexibility and stability.
Pavement Design Question 5:
An existing flexible pavement that develops extensive cracks is called
Answer (Detailed Solution Below)
Pavement Design Question 5 Detailed Solution
Explanation:
Alligator cracks:
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Network of interconnected cracks forming a pattern like alligator skin.
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Caused by repeated traffic loads causing fatigue failure in the pavement layers.
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Indicates loss of structural integrity in the asphalt surface or base layers.
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Often starts as small cracks and spreads over time with traffic.
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Requires major repair such as overlay or complete reconstruction.
Additional Information
Pot hole:
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Circular or irregular holes formed on the pavement surface.
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Caused by water infiltrating through cracks, weakening the base or subgrade.
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Traffic loads cause the weakened area to collapse, forming a hole.
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Can cause damage to vehicles and increase accident risk.
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Usually repaired by cleaning and filling the hole with patching material.
Shear failure:
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Localized pavement failure due to lateral sliding or displacement under heavy loads.
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Occurs often on steep slopes or weak subgrades.
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Manifests as cracks or deformation parallel to the traffic flow.
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Can lead to rutting and surface distress.
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Requires improving subgrade strength or drainage.
Ravelling:
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Gradual loss of surface aggregates from asphalt wearing course.
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Caused by aging binder, poor mixing, or inadequate compaction during construction.
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Results in rough, porous surface with reduced skid resistance.
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Increases susceptibility to moisture damage and further deterioration.
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Often treated by seal coating or surface overlays.
Top Pavement Design MCQ Objective Questions
The interface treatment provided to plug in the voids of porous surfaces and to bond loose particles in bituminous pavements is called:
Answer (Detailed Solution Below)
Pavement Design Question 6 Detailed Solution
Download Solution PDFExplanation:
Prime coat:
The prime coat is an application of low viscous cutback bitumen to an absorbent surface like granular bases on which binder layer is placed. It provides bonding between two layers. Unlike the tack coat, prime coat penetrates into the layer below, plugs the voids, and forms a watertight surface.
Seal Coat:
Seal Coat Seal coat is a thin surface treatment used to water-proof the surface and to provide skid resistance.
Tack coat:
Tack coat is a very light application of asphalt, usually asphalt emulsion diluted with water. It provides proper bonding between two layers of binder course and must be thin, uniformly cover the entire surface, and set very fast.
The bituminous prime coat is the first application of low viscosity liquid bituminous material over an existing porous or absorbent pavement surface like the WBM base course.
Which of the following it NOT a critical parameter to control cracking and rutting in a flexible pavement?
Answer (Detailed Solution Below)
Pavement Design Question 7 Detailed Solution
Download Solution PDFExplanation:
Flexible pavement is modeled as an elastic multilayer structure. Failure criteria in flexible pavement are of following types:
1) Fatigue Cracking
- It is due to the build-up of horizontal tensile strain at the bottom of the asphaltic concrete layer.
- The pavement is considered to fail if 20 % of the surface has cracked.
2) Rutting failure
- It is due to the build-up of excessive compressive strain at the top of the subgrade layer.
- The pavement is considered failed if it exhibits a rut depth of 20 mm.
Note:
According to IRC 37: 2018, Theoretical calculations suggest that the tensile strain near the surface close to the edge of the wheel can be sufficiently large to initiate longitudinal surface cracking followed by transverse cracking much before the flexural cracking of the bottom layer occurs, if the mix tensile strength is not adequate at higher temperatures.
So, Vertical sub-base strain is not a critical parameter to control cracking and rutting in flexible pavement.
Calculate spacing between expansion joints, if the expansion joint gap is 2.0 cm in a cement concrete pavement. The laying temperature is 10°C and the maximum slab temperature in summer is 50°C.(α = 10 × 10-6)
Answer (Detailed Solution Below)
Pavement Design Question 8 Detailed Solution
Download Solution PDFConcept:
Expansion joint:
\(\frac{Δ }{2} = Lα t\)
Where, Δ = width of expansion joint (This gap is such that Δ/2 distance is always maintained after expansion i.e. filler material generally made of cork, Fibre-board is assumed to be compressed by 50%)
L = Length of the slab or spacing between transverse joint
α = Thermal coefficient of concrete per ◦C (Generally taken 10 × 10-6)
t = Temperature difference in ◦C
Calculation:
Δ = 2 cm = 0.02 m
α = 10 × 10-6
t = 50◦C - 10◦C = 40◦C
\(\frac{Δ }{2} = Lα t\)
\(\frac{0.02 }{2} = L× 10 × 10^{-6} × 40\)
L = 25 m
What nature of warping stresses are generated in a reinforced cement concrete pavement during a summer mid-day?
Answer (Detailed Solution Below)
Pavement Design Question 9 Detailed Solution
Download Solution PDFConcept:
The warping stresses are generated in a reinforced cement concrete pavement during a summer mid-day is Tensile in bottom fibre and compressive in top fibre.
Stresses in rigid pavements are divided into 2 categories:
i) Wheel load stress
ii) Temperature stresses
Temperature stresses are classified into 2 categories:
i) Warping stresses: Due to daily variation of temperature
During day time the top surface of the pavement is at a higher temperature as it is directly exposed to the sun than the bottom surface, therefore there occurs a temperature difference between the top and bottom layer. Which causes warping of the pavement.
As at top slab tries to expand to resist that internal compression stress develops on top.
At the bottom, the slab tries to contract but to resist that internal tensile stress develops at the bottom.
During night time the top surface of the pavement is at a lower temperature than the bottom surface, therefore there occurs a temperature difference between the top and bottom layer. Which causes warping of the pavement.
As top slab tries to contract but to resist it internal tensile stresses develop at the top
And compressive stress at the bottom.
ii) Frictional stresses: Due to seasonal variation of temperature.
Change in seasonal temperature induces frictional tension or frictional compression due to expansion and contraction.
In an analysis, frictional stresses are assumed as constant throughout the length lx but actually, frictional stresses are zero as corners and maximum at center.
In winter the slab will try to contract but the frictional resistance between the slab and the soil will resist the slab to contract. Frictional force will develop from both side.
During summer compressive force is induced at top and bottom.
iii) The worst combination of stresses
Position |
Location |
Wheel load |
Warping |
Frictional |
||
|
|
|
Day |
Night |
Summer |
Winter |
Interior |
Top |
C |
C |
T |
C |
T |
bottom |
T |
T |
C |
C |
T |
|
Edge |
top |
C |
C |
T |
C |
T |
bottom |
T |
T |
C |
C |
T |
|
Corner |
top |
T |
C |
T |
|
|
bottom |
C |
T |
C |
|
|
Concrete is weak in tension and therefore a combination of tensile stresses are worst condition.
∴ Concrete slabs (Pavements) are to be designed for these worse combinations.Minimum thickness of the base of a flexible pavement is
Answer (Detailed Solution Below)
Pavement Design Question 10 Detailed Solution
Download Solution PDFConcept:
Flexible pavements:
- Flexible pavements are so named because the total pavement structure deflects, or flexes, under loading.
- Flexible pavement may be constructed in a number of layers and the top layer has to be of the best quality to sustain maximum compressive stress, in addition, to wear and tear.
- The lower layers will experience a lesser magnitude of stress and low-quality material can be used.
- The thickness of the base of a flexible pavement is 10 - 30 cm.
Flexible pavement |
Rigid pavement |
1.Grain to grain load transfer |
1.Slab action takes place |
2.Initial cost is low |
2.Initial cost is high |
3.Joints are not required |
3.Joints are required |
4.Good subgrade is required |
4.Good subgrade is not required |
5.Life of span is short -15 years |
5.Long life span – 30 years |
6.Thickness is More |
6.Thickness is less |
7. Use code IRC 37 |
7. Use code IRC 58 |
The following formula is used to calculate the Equivalent Axle load Factor (EALF) for single axle load (Kg) in vehicle damage factor analysis
Answer (Detailed Solution Below)
Pavement Design Question 11 Detailed Solution
Download Solution PDFConcept:
Vehicle Damage Factor:
- The vehicle damage factor (VDF) is a multiplier to convert the number of commercial vehicles of different axle loads and axle configurations to the number of standard axle load repetitions.
- It is defined as an equivalent number of standard axles per commercial vehicle.
For Single Axle Load:
EALF = \(({axle\: load\over 8160})^4\)
For Tandem Axle Load:
EALF = \(({axle \:load\over 14968})^4\)
Where EALF = Equivalent Axle load Factor
Select the correct option for the given statements.
Statement 1: If CBR for 5 mm exceeds that for 2.5 mm, the CBR test should be repeated.
Statement 2: If identical results follow, the CBR corresponding to 2.5 mm penetration should be taken for design.
Answer (Detailed Solution Below)
Pavement Design Question 12 Detailed Solution
Download Solution PDFConcept:
CBR Test:
- The California bearing ratio test is penetration test meant for the evaluation of subgrade strength of road and pavements. The results obtained by these tests are used with the empirical curves to determine the thickness of pavement and its component layers.
- This is the most widely used method for the design of flexible pavement.
- Generally CBR value is calculated at 2.5 mm and 5 mm penetration.
- The CBR value at 2.5 mm penetration will be greater than the CBR value at 5 mm penetration and in such a case the former will be taken as the CBR value.
- If CBR for 5 mm exceeds that for 2.5 mm, the test should be repeated. If identical result follows, the CBR for 5 mm penetration should be taken for design.
Which of the following is used in a regular pavement maintenance activity?
Answer (Detailed Solution Below)
Pavement Design Question 13 Detailed Solution
Download Solution PDFExplanation:
Fog seal:
(i) Fog seal is used to repair the defects.
(ii) It is a spray of slow setting emulsion diluted with equal amount of water at a rate 0.5 - 1 lit/m2.
(iii) Traffic is allowed after the seal sets in
(iv) It is used to increase the binder content of bituminous surface.
(v) It can also be used as an emergency treatment measure for hungry surface.
Additional Information
Prime coat:
(i) Bituminous prime coat is the first application of a low viscosity liquid bituminous material over an existing porous or absorbent pavements surface like the WBM coarse.
(ii) The main objective of priming is to plug in the capillary voids of the porous surface and to bond the loose mineral particles on the existing surface using a binder of low viscosity which can penetrate into the voids.
Tack coat:
(i) Bituminous tack coat is the application of bituminous material over an existing pavement which is relatively impervious like an existing bituminous surface or cement concrete pavement or a pervious surface like the WBM which has already been treated by a prime coat.
Group index method of designing flexible pavement is based on
a. Plasticity index
b. Shear strength
c. CBR value
d. Percent fines
Answer (Detailed Solution Below)
Pavement Design Question 14 Detailed Solution
Download Solution PDFConcept:
Group Index method:
- It is an empirical method that is based on the physical properties of the soil sub-grade.
- In this method, the thickness of the pavement is decided based on the group index value which is given by -
G.I = 0.2a + 0.005ac + 0.01bd
where,
a - percentage of soil passing through 75μ sieve in excess of 35%, not exceeding 75%
b - percentage soil passing through 75μ sieve in excess of 15%, not exceeding 55%
c - liquid limit in percent in excess of 40% ≯20%
d - plasticity index in excess of 10%≯ 20%
Important Points
- The GI value of soil varies in the range of 0 to 20.
- The following table shows the suitability of soil for subgrade -
Type of soil | GI Value |
Good soil | 0 - 1 |
Fair Soil | 2 - 4 |
Poor Soil | 5 - 9 |
Very poor soil | 10 - 20 |
If the modulus of subgrade reaction of a soil is 20 kg/cu.cm when tested with a 30 cm diameter plate, the corrected modulus of subgrade reaction for the standard diameter plate will be -
Answer (Detailed Solution Below)
Pavement Design Question 15 Detailed Solution
Download Solution PDFConcepts:
Modulus of Subgrade Reaction:
It is defined as the pressure per unit deformation of the subgrade at specific pressure or deformation. It is expressed as:
k = p/Δ
Here,
'k' is the modulus or coefficient of subgrade reaction,
'p' is the pressure, and 'Δ ' is the deformation of soil settlement.
The relation between modulus of sub-grade reaction (k)and plate diameter (d) is given as:
k × d = constant
The standard size plate has dia = 75 cm
Calculation:
Given data;
k1 = 20 kg/cu.cm , d1= 30 cm, and d2 = 75 cm
k2 = the corrected modulus of subgrade reaction for the standard diameter plate
The relation between modulus of sub-grade reaction (k)and plate diameter (d) is given as:
k × d = constant
So
k1 × d1= k2 × d2
20 × 30 = K2 × 75
∴ k2 = 8 kg/cu.cm