Pavement Design Questions and Answers – Highway Maintenance – Benkelman Beam Deflection Studies – 2

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This set of Pavement Design Objective Questions & Answers focuses on “Highway Maintenance – Benkelman Beam Deflection Studies – 2″.

1. The test started from an initial point is stopped after ______ for intermediate deflection measurement.
a) 3 m
b) 2.7 m
c) 8.7 m
d) 9 m
View Answer

Answer: b
Explanation: The Benkelman beam is placed at the first point and the initial deflection is measured. Then it is moved 2.7 m from that point to measure the intermediate deflection. Another 9 m from the second point, the final deflection measurement is taken.
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2. BBD studies were conducted on a pavement at three points and the following data were recorded:

point Do (div) Di (div) Df (div)
A 0 48 45
B 0 34 30
C 5 59 57

Find the rebound deflection at each point i.e. A, B and C respectively if least count of dial gauge is 0.01 mm and the value of k is 2.91.
a) 0.28 mm, 0.63 mm and 1.96 mm
b) 0.82 mm, 0.36 mm and 1.69 mm
c) 1.82 mm, 1.36 mm and 0.69 mm
d) 1.28 mm, 1.63 mm and 0.96 mm
View Answer

Answer: d
Explanation: Before starting the calculations, it is necessary to check for leg correction.
Point A:
Di-Df=48-45=3 div≈0.03 mm
Since Di-Df>0.025 mm, leg correction needs to be applied and rebound deflection can be found using D=0.02(Do-Df )+0.02k(Di-Df ) mm. The value of D0 is 0 and it is to be taken as 100.
D=0.02(100-45)+0.02×2.91(48-45)=1.28 mm
Point B:
Di-Df=34-30=4 div≈0.04 mm
Since Di-Df>0.025 mm, leg correction needs to be applied and rebound deflection can be found using D=0.02(Do-Df )+0.02k(Di-Df ) mm. The value of D0 is 0 and it is to be taken as 100.
D=0.02(100-30)+0.02×2.91(34-30)=1.63 mm
Point C:
Di-Df=59-57=2 div≈0.02 mm
Since Di-Df<0.025 mm, there is no need to apply the leg correction and rebound deflection can be found using D=0.02(Do-Df ) mm. The value of D0 is 5 and it is to be taken as 105.
D=0.02(105-57)=0.96 mm
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3. ______ deflection is obtained as the sum of mean rebound deflection and standard deviation.
a) Specific
b) Characteristic
c) Mechanistic
d) Definite
View Answer

Answer: b
Explanation: Characteristic rebound deflection is used to represent the sum of mean rebound deflection and the standard deviation. The value of standard deviation taken varies with respect to the traffic conditions. For the traffic < 1500 CVPD indicating minor road, one times standard deviation is taken and for > 1500 CVPD indicating major roads, two times the standard deviation is taken.

4. Find the standard deviation for a highway with heavy traffic if the rebound deflection values obtained at 10 points are 1.4, 1.33, 1.36, 1.44, 1.58, 1.62, 1.60, 1.55, 1.47 and 1.38.
a) 0.01
b) 0.11
c) 1.11
d) 1.01
View Answer

Answer: b
Explanation: The equation to find the standard deviation is given by S.D=\(\sqrt{\frac{∑(D_m-D_x)^2}{n-1}}\), where Dm is the mean deflection, Dx is the rebound deflection at each point and n is the number of points. So, therefore, first the difference Dm-Dx is to be found out for each point, it is squared and their sum is taken.
Mean Dm= \(\frac{1.4+1.33+1.36+1.44+1.58+1.62+1.60+1.55+1.47+1.38}{10}\)=1.47
Sum = (1.47-1.4)2+(1.47-1.33)2+(1.47-1.36)2+(1.47-1.44)2+(1.47-1.58)2+(1.47-1.62)2+(1.47-1.60)2+(1.47-1.55)2+(1.47-1.47)2+(1.47-1.38)2
= 0.10
So, S.D=\(\sqrt{\frac{0.10}{10-1}}\)=0.11
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5. What is the standard temperature taken for the rebound deflection calculation?
a) 27°C
b) 30°C
c) 35°C
d) 25°C
View Answer

Answer: c
Explanation: The temperature during the Benkelman beam deflection testing is to be noted. If it is not carried out at the standard temperature of 35°C, a correction for the same is to be provided. If the temperature is higher, the bituminous surface would be soft and deflect more.

6. It is not necessary to apply a correction for temperature if the testing temperature is lower than 35°C.
a) True
b) False
View Answer

Answer: b
Explanation: It is necessary to apply corrections for testing temperatures both lower and higher than 35°C. if the testing temperature is lower, the correction is positive and if it is higher, the correction to be applied is negative.
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7. BBD studies were carried out on a highway pavement at a temperature of 40°C. The rebound deflection was found to be 1.41 mm. what would be the corrected rebound deflection?
a) 1.33 mm
b) 1.35 mm
c) 1.40 mm
d) 1.36 mm
View Answer

Answer: d
Explanation: The correction for temperature is to be applied, as in the question the test temperature is 40°C, but the standard is 35°C. So, the correction to be applied is negative and it is taken as 0.01 per degree.
Corrected Dc=Dc+(Standard temperature-field temperature)×0.01
Corrected Dc=1.41+(35-40)×0.01=1.36 mm

8. The testing is to be carried out during the ______ season.
a) Monsoon
b) Winter
c) Summer
d) Autumn
View Answer

Answer: a
Explanation: The deflection studies should be carried out during the worst possible condition of the pavement. The monsoon season is when this can be achieved. If carried out at other times, then the correction for moisture has to be done.
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9. What would be the corrected rebound deflection if the rebound deflection obtained from BBD studies is 1.33 mm? The test was conducted at 35°C on sandy soil with 5% moisture content. The annual rainfall of the area is obtained as 1050 mm.
a) 1.50 mm
b) 1.56 mm
c) 1.48 mm
d) 1.46 mm
View Answer

Answer: b
Explanation: There is no need to apply temperature correction because the test has been carried out at standard temperature. The correction for moisture content has to be done by referring to the graphs provided in IRC 81:1997. The code can be accessed from the below link:
https://archive.org
The graph labelled as the code can be used for sandy soils with annual rainfall < 1300 mm. From the graph, the correction factor is obtained as 1.17.
Corrected Dc=Dc×correction factor
Corrected Dc=1.33×1.17=1.56 mm

10. What would be the corrected rebound deflection if the rebound deflection obtained from BBD studies is 1.38 mm? The test was conducted at 30°C on a clayey soil having a plasticity of 10% and a moisture content of 4.5%. The annual rainfall of the area is obtained as 1450 mm.
a) 2.57 mm
b) 1.57 mm
c) 2.75 mm
d) 1.75 mm
View Answer

Answer: a
Explanation: The temperature correction needs to be applied first and then the moisture correction on the temperature corrected value.
Corrected Dc=Dc+(Standard temperature-field temperature)×0.01
Corrected Dc=1.38+(35-30)×0.01=1.43 mm
The correction for moisture content has to be done by referring to the graphs provided in IRC 81:1997. The code can be accessed from the below link:
https://archive.org
The graph labelled as in the code can be used for clayey soils with PI < 15% and annual rainfall > 1300 mm. From the graph, the correction factor is obtained as 1.8.
Corrected Dc=Dc×correction factor
Corrected Dc=1.43×1.8=2.57 mm

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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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