# Pavement Design Questions and Answers – Highway Pavements – IRC Method of Design – 1

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

1. What is the method used for the design of rigid pavements?
a) IRC method
b) Westergaard’s analysis
d) PCA method

Explanation: Currently in India, the IRC method is being followed. PCA method is followed by the USA. In the IRC method, both Westergaard’s and Bradbury’s analysis are used in formulating the equations for the design process.

2. The traffic volume is usually projected for ______ years.
a) 15
b) 20
c) 25
d) 30

Explanation: In the equation to compute the traffic volume, the design life is taken as 20 years. A rigid pavement can last up to 30 years, but for design purposes, a period of 20 years is considered.

3. Which of the below axles are not to be considered in the design?
a) Tandem axle
b) Single axle
c) Rear axle
d) Front axle

Explanation: The front axle and rear axle can be single or dual. For design purposes, the front axle is not considered because they carry a lower load and cause less flexural stress on the pavement than the rear axle.
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4. The design of rigid pavement essentially means the computation of slab thickness.
a) True
b) False

Explanation: The design of rigid pavement involves the design of the thickness of the slab, the spacing of joints, design of dowel and tie bar and the design of reinforcement. The slab thickness design is just the first step in the design process.

5. Which of the below is used as a check for the design of slab thickness?
a) Edge temperature stress
b) Corner temperature stress
d) Corner warping stress

Explanation: The slab thickness is designed for edge load stress. After the design is found safe for that, the check is done for temperature stress at the edge and the load stress at the corner. The options corner temperature stress and corner warping stress indicate the same thing.

6. The equation $$f_{cr}=0.49f_{ck}^{0⋅55}$$ to find the flexural strength of concrete was given by ______
a) IS 456:2000
b) Croney and Croney
c) AASHTO
d) IRC 58:2002

Explanation: The equation to $$f_{cr}=0.49f_{ck}^{0⋅55}$$ find the flexural strength of concrete was given by Croney and Croney when gravel aggregates are used. The equation can be slightly modified for crushed aggregates as $$f_{cr}=0.36f_{ck}^{0⋅7}$$ .

7. What would be the design traffic for a National highway design if the following data is available?
Present traffic = 2000 CVPD
Design period = 25 years
Rate of traffic growth = 6%
a) 19626490 CV
b) 4906623 CV
c) 9813245 CV
d) 2453312 CV

Explanation: The equation to find the cumulative repetitions is C=$$\frac{365A[(1+r)^n-1]}{r}$$
Where A is the present traffic = 2000 CVPD
R is the growth rate = 0.006
N is the design life = 25 years
C=$$\frac{365×2000×[(1+0.006)^{25}-1]}{0.006}$$=19626490 commercial vehicles. Now design traffic is to be taken as 25% of C, i.e. the design traffic = 0.25×19626490=4906623 commercial vehicles or CV.

8. What is the range of values for flexural strength of concrete?
a) 44.27 to 47.61 kg/cm2
b) 37.26 to 44.27 kg/cm2
c) 33.27 to 47.61 kg/cm2
d) 37.26 to 47.61 kg/cm2

Explanation: According to IRC 58:2002, there are three distinct values of flexural strength of concrete of grade M40 that are obtained using certain equations. The values are 44.27 kg/cm2 as per IS 456:2000, 37.26 kg/cm2 as per equation by Croney and Croney for gravel and 44.27 kg/cm2 for crushed aggregate. So, the range is from 37.26 to 47.61 kg/cm2.

9. The states belonging to which zone have the maximum temperature differential?
a) Zone I
b) Zone II
c) Zone III
d) Zone IV

Explanation: There is a table given in IRC 58:2002 that provides the values of temperature differentials for various states of India according to the slab thickness. According to that, Zone III includes states like Maharashtra, Karnataka, Andhra Pradesh, etc.… and the maximum temperature differential of 21°C for 30 cm slab has been provided.

10. What is stress ratio?
a) Flexural stress due to load / flexural strength of concrete
b) Flexural strength of concrete / flexural stress due to warping
c) Flexural stress due to warping / flexural strength of concrete
d) Flexural strength of concrete / flexural stress due to load

Explanation: Stress ratio is a term that gives the ratio between the flexural stress due to load and the flexural strength of concrete. It is used in the design process of slab thickness determination. The flexural strength of concrete is usually taken as 45 kg/cm2.

11. To calculate the corner stress, the equation by ______ is used.
a) Westergaard
b) Kelly
c) Teller

Explanation: The equation that is given by Westergaard’s analysis and is modified by Kelly is used to compute the corner stress required to design the pavement. The equation can be written as Sc=$$\frac{3P}{h^2} \left[1-\left(\frac{a√2}{l}\right)^{1⋅2}\right]$$. The equation has the addition of the power of 1.2 to the original equation.

12. The trial thickness is considered to be unsafe if the cumulative fatigue life is ______
a) Less than 1
b) More than 1
c) Less than 0.5
d) More than 0.5

Explanation: As per clause 5.4 in IRC 58:2002, the assumed thickness of the slab is considered to be unsafe if the cumulative fatigue life is found to be more than or equal to one. So, the assumed thickness is no longer valid and another trial value has to be assumed.

13. Which of the below represents a safe design?
a) Temperature stress + load stress at edge = 46.65 kg/cm2
b) Corner load stress = 48 kg/cm2
c) Temperature stress + load stress at edge = 43.65 kg/cm2
d) Corner load stress = 45.65 kg/cm2

Explanation: The temperature and corner load stress are found out as checks to the slab thickness found safe in fatigue life. The values of temperature stress + load stress at the edge and the corner load stress must be less than the flexural strength of concrete i.e. 45 kg/cm2 in order to be safe.

14. The IRC method of rigid pavement design consists of the determination of various parameters. Which of the below options represents the order of designing the parameters?
a) Slab thickness, dowel and tie bar design
b) Joint spacing, slab thickness, dowel and tie bar design
c) Slab thickness, joint spacing, dowel and tie bar design
d) Slab thickness, stress determination, dowel and tie bar design

Explanation: The various parameters are spacing of joints, thickness of slab, dowel bar design, tie bar design and check for the design. In order to compute the stress that is used as a check for determining slab thickness, the spacing of joints has to be known. The designing in order are the spacing of joints, slab thickness, dowel bar and tie bar design.

15. Design the thickness of a rigid pavement slab for a two-way, two-lane National highway in the state of West Bengal. The total two-way traffic is 2500 CVPD at the end of the construction period. The flexural strength of concrete is 45 kg/cm2, modulus of subgrade reaction is 8 kg/cm3, tyre pressure is 8 kg/cm2, the rate of traffic increase is taken as 7.5 %, the spacing of contraction joints is 4.5 m and the width of the slab is 3.5 m. Assume any other data required. The axle load spectrum obtained from a survey is given below:

19-21 0.6 34-38 0.3
17-19 1.5 30-34 0.3
15-17 4.6 26-30 0.6
13-15 11.0 22-26 1.8
11-13 22.3 18-22 1.5
9-11 23.0 14-18 0.5
<9 30.0 <14 2.0

a) 38 cm
b) 36 cm
c) 32 cm
d) 35 cm

Explanation: The procedure is followed exactly as mentioned in the Appendix-2 of IRC 58:2002. The first step is to find the design traffic by evaluating the cumulative repetitions of commercial vehicles.
The cumulative repetitions, C=$$\frac{365A[(1+r)^n-1]}{r}$$
Where A is the present traffic = 2500 CVPD
R is the growth rate = 0.075
N is the design life = 20 years
C=$$\frac{365×2500×[(1+0.075)^{20}-1]}{0.075}$$=39515522 commercial vehicles. Now design traffic is to be taken as 25% of C, i.e. the design traffic = 0.25×39515522=9878881 commercial vehicles.
Now the axle load can be taken as the mid-value of the axle load class and the expected repetitions can be found out by multiplying the respective percentages of load by the design traffic. This can be written in tabular format as below:

Axle load (tons) Expected repetitions Axle class (tons) Expected repetitions
20 59273 36 29637
18 148183 32 29637
16 454429 28 59273
14 1086677 24 177820
12 2202991 20 148183
10 2272143 16 49394
<10 2963664 <16 197578

Now assuming a thickness of 32 cm for the slab and the load factor to be 1.2, the following can be computed and tabulated as below:
The stress from the chart is calculated for slab thickness 32 cm and subgrade reaction modulus of 8 kg/cm3 using the charts in IRC 58:2002 (Refer Book: Guidelines for the Design of Plain Jointed Rigid Pavements Design for Highways, Published by: The Indian Roads Congress).
Stress ratio is obtained as the ratio of flexural stress of concrete i.e. 45 kg/cm2 to the flexural stress due to load i.e. stress from charts.
The fatigue life corresponding to stress ratio is available in Table 6 of IRC 58:2002. It is found out until it becomes infinite.

Axle load (AL) AL X 1.2 Stress from charts Stress ratio Expected repetitions (n) Fatigue life (N) Fatigue life consumed = n/N
Single axle
20 24.0 25.19 0.56 59273 94.1 x 103 0.63
18 21.6 22.98 0.51 148183 4.85 x 105 0.31
16 19.2 20.73 0.46 454429 14.33 x 106 0.03
14 16.8 18.45 0.41 1086677 Infinite 0
Tandem axle
36 43.2 20.07 0.45 29637 62.8 x 106 0.004
32 38.4 18.40 0.40 29637 Infinite 0
Cumulative fatigue life consumed = 0.9704

Since the cumulative fatigue life consumed is less than 1, the thickness assumed is safe.
The next step is to check for temperature stress and corner stress.
Temperature stress at the edge is to be checked for using the equation $$St_e=\frac{CEet}{2}$$. Finding the radius of relative stiffness as below:
Assuming E = 3×105 kg/cm2 and μ as 0.15,
l=$$\left[\frac{Eh^3}{12k(1-μ^2)}\right]^{1∕4}=\left[\frac{3×10^5×32^3}{12×8×(1-0.15^2)}\right]^{1∕4}$$=101.17 cm
L/l ratio for computing C = 450/101.17 = 4.5. Using Figure 2 from IRC 58:2002, the value of C is obtained as 0.58; e is assumed as 10×10-6 /°C and t for West Bengal can be obtained as 16.8°C from Table 1 given in IRC 58:2002.
$$St_e=\frac{0.58×3×10^5×10×10^{-6}×16.8}{2}$$=14.62 kg/cm2
The stress = temperature stress + maximum load stress = 14.62 + 24 = 38.62 kg/cm2, which is less than the flexural stress of concrete (45 kg/cm2), therefore the design is correct.
The equation to find corner stress is Sc=$$\frac{3P}{h^2} \left[1-\left(\frac{a√2}{l}\right)^{1⋅2}\right]$$. The design load P is the 98-percentile axle load = 16 tonnes, so the wheel load is 8 tonnes or 8000 kg. the tyre pressure, p is given as 8 kg/cm2. The radius of the contact area, a is found using the relation, $$\frac{π}{4}a^2=\frac{P}{p}$$ which can be rewritten as a=$$\sqrt{\frac{4P}{πp}}=\sqrt{\frac{4×8000}{π×8}}$$=35.68 cm.
Therefore, Sc=$$\frac{3×8000}{32^2} \left[1-\left(\frac{35.68√2}{101.17}\right)^{1⋅2}\right]$$ =13.27 kg/cm2. The corner stress 13.27 kg/cm2 is less than the flexural stress of concrete (45 kg/cm2), therefore the design is correct.
The slab thickness is designed as 32 cm.

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