Pavement Design Questions and Answers – Highway Pavements – IRC Method of Design – 2

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This set of Pavement Design Questions & Answers for Exams focuses on “Highway Pavements – IRC Method of Design – 2”.

1. Which of the below recommendations for the parameters used in the design is correct?
a) The minimum value of k is 5 kg/cm3
b) The flexural strength of cement should be more than 40 kg/cm2
c) The design wheel load is taken as 5100 kg
d) The modulus of elasticity is taken as 2×105 kg/cm2
View Answer

Answer: c
Explanation: As a general rule, some specific considerations are used in the design of rigid pavements. The minimum value of k is 5.5 kg/cm3 , the flexural strength of cement should be less than 40 kg/cm2 in order to be safe and the modulus of elasticity is taken as 3×105 kg/cm2. The design wheel load is 5100 kg.

2. If the total repetitions of commercial vehicles are known, then the design traffic is taken as ______ of it.
a) 2 times
b) 25 %
c) Half
d) 50 %
View Answer

Answer: b
Explanation: The design traffic is taken as 25 % of the total repetitions of commercial vehicles. This criterion has been mentioned in the details under clause 4.4 design traffic of IRC 58:2002. Half and 50 % are the same.

3. The temperature differential for different states is expressed in a tabular format. What are the various slab thicknesses for which the same is given?
a) 10, 20, 30 cm
b) 10, 15, 20, 25, 30 cm
c) 15, 20, 25, 30 cm
d) 15, 25, 35 cm
View Answer

Answer: c
Explanation: The temperature differential has been given for varying thickness of slab as well as different states of India. According to IRC 58:1988, the thickness ranging from 10, 15, 20, 25 and 30 cm were given. But in the latest one, the range is 15, 20, 25 and 30 cm.
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4. Which of the below institute and the value of modulus of elasticity given by them are not matched correctly for M40 grade concrete?
a) IS 456:2000 – 31623 N/mm2
b) ACI building code:31889 – 32000 N/mm2
c) BS:8110 (Part 2) – 28000 N/mm2
d) PCA – 25000 N/mm2
View Answer

Answer: d
Explanation: There are various values recommended by different institutes for concrete of grade M40. The recommended value by PCA is 28000 N/mm2 and not 25000 N/mm2. Generally, the range of values for modulus of elasticity have found to be in the range of 30741 to 31623 N/mm2.

5. Find the flexural strength of concrete of M40 grade using the equation given by Croney and Croney if crushed aggregates are used in the design of pavement.
a) 47.62 kg/cm2
b) 48 kg/cm2
c) 47.16 kg/cm2
d) 47 kg/cm2
View Answer

Answer: a
Explanation: The equation for crushed aggregates by Croney and Croney is given by \(f_{cr}=0.36f_{ck}^{0⋅7}\). Substituting the value of fck as 400 kg/cm2, the value of flexural strength is obtained as below:
fcr=0.36×4000.7=47.62 kg/cm2.

6. The fatigue life is taken as unlimited if the value of stress ratio is ______
a) < 0.55
b) > 0.45
c) < 0.45
d) > 0.55
View Answer

Answer: c
Explanation: The fatigue life and the stress ratio can be related to each other using different equations for the range of stress ratio. When the stress ratio is less than 0.45, the fatigue life is considered to be unlimited. For the condition 0.45 ≤stress ratio≤ 0.55, there is an equation and for stress ratio > 0.55, there is another equation.

7. Design a tie bar if the trial thickness of 30 cm was found safe. The other details are:
Width of the lane = 3.5 m.
Allowable working stress in concrete = 1250 kg/cm2
Unit weight of concrete = 2400 kg/m3
Coefficient of friction = 1.5
Allowable bond stress in plain bars = 17.5 kg/cm2
Allowable tensile stress in plain bars = 1250 kg/cm2.
a) 14 mm diameter bar of 58 cm length placed at 38 cm spacing
b) 12 mm diameter bar of 58 cm length placed at 38 cm spacing
c) 14 mm diameter bar of 60 cm length placed at 40 cm spacing
d) 12 mm diameter bar of 60 cm length placed at 40 cm spacing
View Answer

Answer: b
Explanation: There are various steps to design the tie bar:
Step 1: Area of steel per m width of joint
\(A_s=\frac{fWbh×10^{-2}}{s_s}\)
Where AS is the area of steel per metre of joint width to resist the frictional force in the slab
f is the coefficient of friction = 1.5
W is the unit weight of concrete = 2400 kg/m3
b is the width of the slab = 3.5 m
h is the thickness of the slab = 30 cm
SS is the allowable tensile stress in plain bars = 1250 kg/cm2
Substituting these in the equation,
\(A_s=\frac{1.5×2400×3.5×30×10^{-2}}{1250}\)=3.02 cm2/m
Step 2: Spacing of tie bar
Assuming the diameter of the plain bar as 1.2 cm, the cross-sectional area can be found as:
\(A=\frac{π}{4}×d^2=\frac{π}{4}×1.2^2=1.13 cm^2\)
The spacing of tie bar = \(\frac{A}{A_s} =\frac{1.13}{3.02}\)=0.3742 m≈37.42 cm
Therefore, the spacing to be provided is 38 cm.
Step 3: Length of the tie bar
L=\(\frac{2S_s A}{S_b P}\)
Where L is the length of the tie bar
SS is the allowable tensile stress in plain bars = 1250 kg/cm2
A is the cross-sectional area of tie bar = 1.13 cm2
P is the perimeter of the tie bar = πd=π×1.2=3.77 cm
Sb is the allowable bond stress in plain bars = 17.5 kg/cm2
After substituting these in the equation,
\(L_t=\frac{2×1250×1.13}{17.5×3.77}\)=42.82 cm
The obtained length has to be increased by 10 cm to account for the loss of bond due to painting and an additional 5 cm as tolerance in placement. Therefore, the length of the tie bar would be
L = 42.82 + 10 + 5 = 57.82 ≈58 cm
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8. The design of dowel bar was carried out and the following were found out:
The load carried by outer dowel bar, Pt=1432 kg
The modulus of dowel/concrete interaction, K = 41500 kg/cm2/ cm
The modulus of elasticity, E = 2×106 kg/cm2
The moment of inertia of dowel bar, I=5.147 cm4
The relative stiffness of bar embedded in concrete, β=0.238
The joint width, Z = 2 cm
Determine the bearing stress in the dowel bar and ascertain its safety.
a) 265 kg/cm2 and safe
b) 265 kg/cm2 and not safe
c) 256 kg/cm2 and safe
d) 256 kg/cm2 and not safe
View Answer

Answer: a
Explanation: The equation to find the bearing stress is Bearing stress=\((P_t k)\frac{2+βz}{4β^3 EI}\) . The terms used in the equation are mentioned in the question and can be substituted directly.
Bearing stress=\(1432×41500×\frac{2+0.238×2}{4×0.238^3×2×10^6×5.147}=265.07≈\)265 kg/cm2
The bearing stress in dowel bar (265 kg/cm2) is found to be less than the bearing stress in concrete (292 kg/cm2), therefore the design is safe.

9. The axle load and the stress ratio of the various axle loads are represented in the table below. The expected number of repetitions is also provided against each axle load. Using the data given below, find the cumulative fatigue life.

Axle load (AL) Stress ratio Expected repetitions (n)
Single axle
20 0.58 67544
18 0.50 165645
16 0.43 487907
Tandem axle
36 0.48 35765
32 0.42 35765
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a) 0.2524
b) 0.2245
c) 0.2445
d) 0.2425
View Answer

Answer: d
Explanation: The fatigue life corresponding to the stress ratios provided in the question can be found out from Table 6 in IRC 58:2002 or using the equations as below.
For stress ratio < 0.45, fatigue life, N is infinite.
For 0.45 ≤stress ratio≤ 0.55, N=\(\left[\frac{4.2577}{SR-0.4325}\right]^{3.268}\)
For stress ratio > 0.55, log⁡N= \(\frac{0.9718-SR}{0.0828}\)
After computing the respective values of fatigue life, they can be written in the table as shown below. The fatigue life consumed is the ratio of expected repetitions and fatigue life.

Axle load (AL) Stress ratio Expected repetitions (n) Fatigue life (N) Fatigue life consumed n/N
Single axle
20 0.52 67544 3.26 X 105

0.21
18 0.47 165645 5.2 X 106

0.03
16 0.42 487907 Infinite 0
Tandem axle
36 0.46 35765 1.4353 X 107

0.0025
32 0.41 35765 Infinite 0

The cumulative fatigue life consumed = 0.21 + 0.03 + 0.0025 = 0.2425.

10. Flexural strength of concrete forms the criterion to evaluate the strength of pavement concrete.
a) True
b) False
View Answer

Answer: a
Explanation: The flexural strength of concrete is related to the compressive strength of concrete. This relation depends on the nature of aggregates, cement type, additives added and other factors. It is evaluated using the flexural test and due to the above reasons, the flexural strength of concrete is considered as a criterion to evaluate the strength of pavement concrete.

11. What is the fatigue life of the pavement if the flexural stress due to load is 18.5 kg/cm2?
a) 6.279 X 107
b) Unlimited
c) 1099
d) 3.34 X 103
View Answer

Answer: b
Explanation: The fatigue life is computed based on the stress ratio, which is flexural stress due to load by flexural strength of concrete. Stress ratio can hence be written as \(\frac{18.5}{45}\)=0.41. The stress ratio is less than 0.45, which means that the fatigue life of the pavement is unlimited.

12. The 7 days unconfined compressive strength of cement-treated granular soil should be ______
a) Minimum of 7 MPa
b) Minimum of 2.1 MPa
c) Maximum of 7 MPa
d) Maximum of 2.1 MPa
View Answer

Answer: b
Explanation: The 7 days unconfined compressive strength of cement-treated granular soil should be a minimum of 2.1 MPa. It is supposed to be a minimum of 7 MPa for dry lean concrete sub-base. This has been recommended and mentioned in clause 4.6.1 in IRC 58:2002.

13. What is the name of the computer programme developed by IIT Kharagpur for computing the edge load stress?
a) IITPAVE
b) IITRIG
c) IITRGD
d) IITRIGID
View Answer

Answer: d
Explanation: The computer programme developed by IIT Kharagpur for the computation of edge load stress was named as IITRIGID. IITPAVE is a programme developed by IIT Kharagpur for the analysis of pavements. The process has been discussed in IRC 37:2012 and later versions.

14. Reinforcement is provided in the slabs to improve their flexural strength.
a) True
b) False
View Answer

Answer: b
Explanation: Reinforcement is provided in the slabs to counteract the tensile stresses and not to improve the flexural strength. Tensile stresses are caused due to the shrinkage or contraction of slabs during temperature or moisture changes.

15. Find the fatigue life of the pavement for a stress ratio of 0.56.
a) 9.41
b) 9.41×104
c) 9.41×105
d) 9.41×102
View Answer

Answer: b
Explanation: The stress ratio is greater than 0.55, so the below equation is to be used to compute the fatigue life.
log⁡N= \(\frac{0.9718-SR}{0.0828}\)
SR = 0.56, so the N is obtained as
N⁡=antilog \(\frac{0.9718-0.56}{0.0828}\)
N=antilog 4.973=94065.41&9.41×104

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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