MCQOPTIONS
Saved Bookmarks
MCQOPTIONS
This section includes 15 Mcqs, each offering curated multiple-choice questions to sharpen your Pavement Design knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Which temperature stress is generally ignored in the design? |
| A. | Frictional stress |
| B. | Interior stress |
| C. | Edge stress |
| D. | Corner stress |
| Answer» E. | |
| 2. |
What would be the frictional stress if the spacing between contraction joints is 5 m, the coefficient of friction is 1.5 and the unit weight of concrete is 2400 kg/m3? |
| A. | 0.9 kg/cm2 |
| B. | 0.88 kg/cm2 |
| C. | 0.8 kg/cm2 |
| D. | 0.99 kg/cm2 |
| Answer» B. 0.88 kg/cm2 | |
| 3. |
Determine the Bradbury coefficients of a 25 cm thick pavement if the following data is provided.Spacing of transverse joint = 10 mSpacing of longitudinal joint = 4 mModulus of subgrade reaction = 7 kg/cm3Radius of contact area = 15 cmPoisson’s ratio = 0.15Modulus of elasticity = 3×105 kg/cm2 |
| A. | Cx=1.05 and Cy=0.64 |
| B. | Cx=0.64 and Cy=1.05 |
| C. | Cx=1.50 and Cy=0.46 |
| D. | Cx=0.46 and Cy=1.50 |
| Answer» B. Cx=0.64 and Cy=1.05 | |
| 4. |
How was the temperature stress analysis done by J. Thomlinson? |
| A. | Thermometers |
| B. | Thermocouples |
| C. | Thermostats |
| D. | Thermocline |
| Answer» C. Thermostats | |
| 5. |
Find the warping stress in the interior of the pavement for the data given below.Temperature differential, t = 17°CPoisson’s ratio, μ = 0.15Modulus of elasticity, E = 2×105 kg/cm2Thermal coefficient, e = 10×10-6 /°CCx=0.75 and Cy=0.56. |
| A. | 15.5 kg/cm2 |
| B. | 14 kg/cm2 |
| C. | 15 kg/cm2 |
| D. | 14.5 kg/cm2 |
| Answer» E. | |
| 6. |
The temperature differential is affected by the geographical features of the pavement location. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 7. |
Find the warping stress at the corner of the pavement for the data given below.Pavement thickness = 25 cmRadius of relative stiffness = 64.50 cmTemperature differential = 0.6°C per cm of slab thicknessRadius of contact area = 15 cmPoisson’s ratio = 0.15Modulus of elasticity = 2.1×105 kg/cm2Thermal coefficient = 10×10-6/°C |
| A. | 6.59 kg/cm2 |
| B. | 5.96 kg/cm2 |
| C. | 6.95 kg/cm2 |
| D. | 5.69 kg/cm2 |
| Answer» C. 6.95 kg/cm2 | |
| 8. |
What would be the frictional stress if the spacing between contraction joints is 4.2 m, the coefficient of friction is 1.3 and the unit weight of concrete is 2400 kg/m3? |
| A. | 0.66 kg/cm2 |
| B. | 0.70 kg/cm2 |
| C. | 0.76 kg/cm2 |
| D. | 0.60 kg/cm2 |
| Answer» B. 0.70 kg/cm2 | |
| 9. |
Determine the warping stress at the edge of a 20 cm thick pavement if the following data is provided.Spacing of transverse joint = 10 mSpacing of longitudinal joint = 3.6 mModulus of subgrade reaction = 6 kg/cm3Temperature differential = 0.5°C per cm of slab thicknessRadius of contact area = 15 cmPoisson’s ratio = 0.15Modulus of elasticity = 3×cm5 kg/cm2Thermal coefficient = 10 ×10-6 /°C |
| A. | 16.6 kg/cm2 |
| B. | 15.6 kg/cm2 |
| C. | 15.5 kg/cm2 |
| D. | 16.5 kg/cm2 |
| Answer» C. 15.5 kg/cm2 | |
| 10. |
Frictional stresses are developed due to ______ |
| A. | Relative movement of the base |
| B. | Relative movement of the slab |
| C. | Daily temperature variation |
| D. | Seasonal temperature variation |
| Answer» E. | |
| 11. |
What is the equation used to compute the frictional stresses in the rigid pavement? |
| A. | \(S_f=\frac{Lf}{2×10^4}\) |
| B. | \(S_f=\frac{wf}{2×10^4}\) |
| C. | \(S_f=\frac{wLf}{2×10^4}\) |
| D. | \(S_f=\frac{wLf}{2×10^5}\) |
| Answer» D. \(S_f=\frac{wLf}{2×10^5}\) | |
| 12. |
When does the warping of the cement concrete slab occur? |
| A. | Different temperature on layers |
| B. | Temperature differential exceeds 30 |
| C. | Same temperature on layers |
| D. | Temperature differential is below 10 |
| Answer» B. Temperature differential exceeds 30 | |
| 13. |
What are the type of stresses induced due to the temperature change in the pavement? |
| A. | Frictional stress and thermal stress |
| B. | Warping stress and thermal stress |
| C. | Warping stress and frictional stress |
| D. | Thermal stress and temporal stress |
| Answer» D. Thermal stress and temporal stress | |
| 14. |
The seasonal variation of temperature causes the variation in temperature across the depth of the slab. |
| A. | True |
| B. | False |
| Answer» C. | |
| 15. |
Temperature stresses in the pavement are caused due to the variation of temperature in ______ |
| A. | Slab |
| B. | Cement |
| C. | Subgrade |
| D. | Sub-base |
| Answer» B. Cement | |