# Pavement Design Questions and Answers – Highway Pavements Design – California Resistance Value Method

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

1. ______ is used to test the resistance value or R-value of the material?
a) Cohesiometer
b) Stabilometer
c) Penetrometer
d) Tribometer

Explanation: The apparatus used to test the resistance of the material is the stabilometer. It gives the R-value of the material. Cohesiometer gives the C-value, penetrometer measures penetration and tribometer is used to measure friction.

2. How is R-value related to C-value?
a) R=C5
b) R=1/C5
c) R=C5
d) R=1/C1/5

Explanation: R-value is the resistance value obtained from the stabilometer test and C-value is the cohesiometer value. Both terms are used for the computation of pavement thickness. R-value is found to vary inversely with the fifth root of C-value.

3. What is the equation to find the thickness of pavement using the R-value?
a) $$T=\frac{K(TI)(90-R)}{C^{1⁄5}}$$
b) $$T=\frac{K(TI)(180-R)}{C^5}$$
c) $$T=\frac{K(TI)(180-R)}{C^{1⁄5}}$$
d) $$T=\frac{K(TI)(90-R)}{C^5}$$

Explanation: $$T=\frac{K(TI)(90-R)}{C^{\frac{1}{5}}}$$ is the right equation used to compute the thickness of the pavement using the California Resistance Value method of design. In the equation, the term K is a constant, TI is the traffic index, R is the resistance value and C is the cohesiometer value.

4. What is the typical R-value for a well-graded crushed stone base course?
a) > 50
b) < 50
c) > 80
d) < 80

Explanation: California resistance value or the R-value is obtained as per AASHTO T 190 and ASTM D 2844. The typical R-value for a well-graded crushed stone base course is greater than 80 and for MH silts, it is in between 15 and 30.

5. A vehicle has four axles, what would be the equivalent wheel load (EWL) constant used for the R-value design method?
a) 4620
b) 330
c) 2460
d) 3040

Explanation: In order to find the traffic index, it is necessary to know the EWL. A Statewide loadometer survey was carried out in California in 1955 and based on the same, EWL constants for various number of axles where obtained. For a vehicle having four axles, the EWL constant is 2460.

6. Calculate the fifteen-year traffic index for the data given below.

 Number of axles Annual Average Daily Traffic AADT (2 directions) 2 3300 3 400 4 278 5 90

Assume 50% increase in traffic in the fifteen-year period.
a) 9.500
b) 9.469
c) 9.496
d) 9.515

Explanation: The EWL for the above data can be tabulated as shown in the table below. The EWL constants are already defined by the survey conducted by California State. EWL is obtained by multiplying AADT and EWL constants.

 No. of axles AADT EWL constants Product 2 3300 330 1089000 3 400 1070 428000 4 278 2460 683880 5 90 4620 415800 Total yearly EWL = 2616680

Taking average increase for the fifteen-year period,
EWL = $$(\frac{1+1.5}{2})$$×15×2616680=49062750
TI = 1.35 EWL0.11=1.35×(49062750)0.11=9.469

7. In order to find the thickness of the pavement, it is necessary to find the equivalent C value of the pavement materials.
a) True
b) False

Explanation: The cohesiometer value or the C-value is obtained for various layers of the pavement after the test. It is difficult to have different C values for the pavement thickness calculations. Therefore, it is necessary to find the equivalent C-value.

8. Calculate the equivalent C-value for a pavement having the details as mentioned in the table below.

 Material Thickness (cm) C-value Bituminous concrete 10 62 Cement treated base 20 200 Gravel sub-base 10 15

a) 87
b) 88
c) 89
d) 86

Explanation: The relation to find the equivalent C-value is $$\frac{T_1}{T_2}=(\frac{C_2}{C_1})^{1/5}.$$ For various layers, this relation has to be applied. T represents the thickness and C represents the C-vale. First, the thickness of each layer is converted in terms of their respective gravel equivalent (g).
For bituminous concrete represented as bc, $$\frac{T_g}{T_{bc}}=(\frac{C_{bc}}{C_g})^{1/5}$$
$$T_g=T_bc (\frac{C_{bc}}{C_g})^{1/5}=10×(\frac{62}{15})^{1/5}=13.28 cm$$
For cement-treated base represented as ctb,$$\frac{T_g}{T_{ctb}} =(\frac{C_{ctb}}{C_g})^{1/5}$$
$$T_g=T_{ctb}(\frac{C_{ctb}}{C_g})^{1/5}=20×(\frac{200}{15})^{1/5}=33.58 cm$$
For gravel sub-base, the thickness Tg=10 cm as in the question.
The actual pavement thickness, T = 10+20+10 = 40 cm
Equivalent to gravel thickness = 13.28+33.58+10 = 56.86 cm
Therefore, the equivalent C-value can be found as
$$\frac{T_g}{T}=(\frac{C}{C_g})^{1/5}$$
$$C=(\frac{T_g}{T})^5×C_g=(\frac{56.86}{40})^5×15=87.06≈87$$

9. Find the thickness of the pavement using R-value method for the following details.
EWL = 36,345,233
R = 42
C = 88
a) 22.98 cm
b) 22.89 cm
c) 29.82 cm
d) 28.92 cm

Explanation: The equation T=$$\frac{K(TI)(90-R)}{C^{1⁄5}}$$ can be used to find the thickness of pavement.
The constant k is already defined as 0.166
Traffic index, TI = 1.35 EWL0.11=1.35×(36345233)0.11=9.162
Therefore, T=$$\frac{0.166×9.162×(90-42)}{88^{1⁄5}}$$ =29.82 cm

10. The C-value of a gravel base course ranges from 22 to 30.
a) True
b) False

Explanation: The typical C-value for various materials have been defined. For a gravel base course, the C-value has been taken as 15. The range of C-value from 22 to 30 is taken for the open-graded bituminous mix.

11. What is the rate of exudation pressure applied on the subgrade for the stabilometer test?
a) 900 kg per minute
b) 1000 kg per minute
c) 950 kg per minute
d) 1050 kg per minute

Explanation: The subgrade is subjected to a pressure to force out the water from the compacted subgrade. This pressure is known as exudation pressure and the rate of the application is 900 kg per minute. It depends on the soil type and moisture content.

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