This set of Pavement Design Multiple Choice Questions & Answers (MCQs) focuses on “Combined Stresses in Rigid Pavement”.

1. What happens to the top of the pavement during summer?

a) Expands

b) Contracts

c) Cracks

d) Shrinks

View Answer

Explanation: During summer, the cement concrete slab on top of the pavement expands towards the expansion joints. The slab gets heated up and the heat leads to the expansion of the slab. The slab contracts during winter. When the expansion and contraction effects increase, cracks are developed.

2. Where is the maximum tensile stress developed during mid-day?

a) Top fibre

b) Bottom fibre

c) Center fibre

d) Side fibre

View Answer

Explanation: During mid-day, the top of the pavement is already heated up more than that of the layer below. Or in other words, the top fibre of the slab is at a temperature higher than that of the bottom fibre. So, the slab tends to warp downwards, hence the maximum tensile stress is developed in the bottom fibre.

3. In the pavement corner regions, the critical stress occurs at ______ fibre during ______

a) Top, mid-day

b) Top, midnight

c) Bottom, mid-day

d) Bottom, midnight

View Answer

Explanation: The critical stress in the corner region of the pavement occurs during midnight. During midnight, the temperature of the bottom fibre is higher than that of the top fibre, therefore the slab tends to warp upwards. So, the critical combination occurs at top fibre during midnight.

4. What is the critical combination of the stresses at the edge of the pavement during winter?

a) Load stress – warping stress

b) Load stress + warping stress

c) Load stress + warping stress + frictional stress

d) Load stress + warping stress – frictional stress

View Answer

Explanation: Load stress + warping stress + frictional stress is the critical combination of the stresses at the edge of the pavement during winter. During winter, the slab contracts giving rise to warping stress and tensile frictional stress. The slab has a lower temperature and the stress due to wheel load has to be considered.

5. What is the critical stress combination at the corners?

a) Load stress + warping stress

b) Frictional stress + warping stress

c) Load stress + frictional stress

d) Load stress + warping stress + frictional stress

View Answer

Explanation: In the corner region of the pavement slab, the critical combination of the stress is given by load stress + warping stress. There is no frictional stress in the case of corner regions because the corners are free to warp.

6. In summer season, the frictional stress is ______

a) Higher

b) Tensile

c) Compressive

d) Lower

View Answer

Explanation: In the summer season, the concrete slab expands. This expansion is resisted by the effect of friction and hence the frictional stress is induced. Frictional stress is in the opposite direction to expansion, i.e. it contracts or compresses the slab, hence the stress is compressional frictional stress.

7. The critical combination of stress occurs at the interior of the pavement during summer.

a) True

b) False

View Answer

Explanation: During summer, the interior as well as the edge of the pavement, are prone to a critical combination of stresses. But, the load stress at the interior is less than that at the edge. Therefore, the critical combination is taken at the edge only.

8. What will be the critical combination of stresses at the corner if the following stresses have been calculated for the pavement?

Load stress = 30.50 kg/cm^{2}

Warping stress = 8.56 kg/cm^{2}

Frictional stress = 0.77 kg/cm^{2}

a) 39.06 kg/cm^{2}

b) 39.83 kg/cm^{2}

c) 31.27 kg/cm^{2}

d) 38.29 kg/cm^{2}

View Answer

Explanation: The critical combination of the stresses at the corner of the pavement can be obtained as the sum of load stress and warping stress. Frictional stress is negligible and hence ignored. Therefore, the critical combination of stresses at the corner = 30.50 + 8.56 = 39.06 kg/cm

^{2}.

9. A cc slab of thickness 20 cm is constructed over a base of granular material with a modulus of reaction 10 kg/cm^{3}. The maximum temperature differential between the top and bottom of the slab is found to be 19°C during summer day and night. The spacing between transverse joints is 8 m and the spacing between longitudinal joints is 4 m. The design wheel load is 5100 kg, radius of contact area is 15 cm, Poisson’s ratio is 0.15, modulus of elasticity is 3×10^{5} kg/cm^{2}, the thermal coefficient is 10 ×10^{-6} /°C and the coefficient of friction is 1.5. Using the above information, find the worst combination of stresses at the edge of the pavement.

a) 51.59 kg/cm^{2}

b) 59.51 kg/cm^{2}

c) 59.15 kg/cm^{2}

d) 51.95 kg/cm^{2}

View Answer

Explanation: The worst combination of the stresses during summer mid-day at the edge of the pavement can be obtained as load stress + warping stress + frictional stress. Each of the stresses has to be found out separately.

The load stress at the edge is given by \(S_e=\frac{0.572P}{h^2}\)[4 logl/b+0.359].

l=\(\left[\frac{Eh^3}{12k(1-μ^2)}\right]^{1∕4}=\left[\frac{3×10^5×20^3}{12×10×(1-0.15^2)}\right]^{1∕4}\)=67.26 cm

a < 1.724h i.e. 15 < 1.724 X 20, therefore,

b=\(\sqrt{1.6a^2+h^2}-0.675h=\sqrt{1.6×15^2+20^2}\)-0.675×20=14.07 cm

Now,

\(S_e=\frac{0.572P}{h^2}[4 logl/b+0.359] =\frac{0.572×5100}{20^2}\) [4 log67.26/14.07+0.359]=22.44 kg/cm

^{2}

The warping stress at the edge is given by \(St_e=\frac{C_x Eet}{2}\) and \(St_e=\frac{C_y Eet}{2}\). The terms C

_{x}and C

_{y}are called Bradbury’s coefficients and their values can be obtained using values given in IRC 58:2002. To find C

_{x}and C

_{y}, the ratios \(\frac{L_x}{l}\) for C

_{x}and \(\frac{L_y}{l}\) for C

_{y}have to be found out. L

_{x}is the spacing between the transverse joints = 800 cm and L

_{y}is the spacing between longitudinal joints = 400 cm.

\(\frac{L_x}{l}\)=800/67.26=11.89 and \(\frac{L_y}{l}\)=400/67.26=5.95

Referring to the values in the chart for determining coefficient C in IRC 58:2002, the interpolation of the exact values can be done as shown below.

L/l | C |
---|---|

5 | 0.72 |

6 | 0.92 |

11 | 1.050 |

12 | 1.00 |

Therefore, for \(\frac{L_x}{l}\)= 11.89, the value would be \(\frac{C_x-1.05}{11.89-11}=\frac{1-1.05}{12-11}\), C_{x}=1.01 and for \(\frac{L_y}{l}\)= 5.95, the value would be \(\frac{C_y-0.72}{5.95-5}=\frac{0.92-0.72}{6-5}\), C_{y}=0.91.

C_{x} has a higher value, therefore, the equation giving the highest stress value \(St_e=\frac{C_x Eet}{2}\) has to be used.

\(St_e=\frac{C_x Eet}{2}=\frac{1.01×3×10^5×10×10^{-6}×19}{2}\)=28.79 kg/cm^{2}

Next step is to compute the frictional stress using the equation \(S_f=\frac{WLf}{2×10^4}\). Here the value of W can be assumed as 2400 kg/m^{3}.

\(S_f=\frac{2400×4×1.5}{2×10^4}\)=0.72 kg/cm^{2}

Therefore, the worst stress combination would be S_{e}+St_{e}+S_{f} = 22.44 + 28.79 + 0.72 = 51.95 kg/cm^{2}

10. The load stress is the least in which region?

a) Corner

b) Edge

c) Interior

d) Bottom

View Answer

Explanation: The load stress decreases in the order at the corner, edge and interior part of the pavement. The temperature stress, on the other hand, increases in that order. Therefore, the corner of the pavement experiences the least stress due to wheel loads.

11. What will be the critical combination of stresses at the edge during winter if the following stresses have been calculated for the pavement?

Load stress = 35.67 kg/cm^{2}

Warping stress = 30.31 kg/cm^{2}

Frictional stress = 0.89 kg/cm^{2}

a) 65.09 kg/cm^{2}

b) 65.90 kg/cm^{2}

c) 69.50 kg/cm^{2}

d) 69.05 kg/cm^{2}

View Answer

Explanation: The critical combination of the stresses at the edge of the pavement during winter can be obtained as load stress + warping stress – frictional stress. Therefore, the critical combination of stresses = 35.67 + 30.31 – 0.89 = 65.09 kg/cm

^{2}.

12. What stresses are to be computed if critical combination during day time is to be found out?

a) Corner stresses

b) Edge stresses

c) Load stresses

d) Warping stresses

View Answer

Explanation: During the day time, the critical combination of load, warping and frictional stresses in the edge region have to be found out. The variation in frictional stress occurs when it is summer (compressive) or winter (tensile). During night time, the critical combination of load and warping stresses at the corner is considered, frictional stresses are neglected at the corners.

13. A cc slab of thickness 25 cm is constructed over a base of granular material with a modulus of reaction 8 kg/cm^{3}. The maximum temperature differential between the top and bottom of the slab is found to be 20°C during summer day and night. The spacing between transverse joints is 4.5 m and the spacing between longitudinal joints is 3.5 m. The design wheel load is 5100 kg, radius of contact area is 15 cm, Poisson’s ratio is 0.13, modulus of elasticity is 2×10^{5} kg/cm^{2}, the thermal coefficient is 10 ×10^{-6} /°C and the coefficient of friction is 1.5. Using the above information, find the worst combination of stresses during mid-night.

a) 16 kg/cm^{2}

b) 15.99 kg/cm^{2}

c) 16.99 kg/cm^{2}

d) 15 kg/cm^{2}

View Answer

Explanation: During mid-night, the worst combination of stresses occurs at the corner of the pavement. It can be obtained as load stress + warping stress.

The load stress is given by \(S_c=\frac{0.316P}{h^2}\)[4 logl/b+1.069].

l=\(\left[\frac{Eh^3}{12k(1-μ^2)}\right]^{1∕4}=\left[\frac{2×10^5×25^3}{12×8×(1-0.13^2)}\right]^{1∕4}\)=75.86 cm

a < 1.724h i.e. 15 < 1.724 X 25, therefore,

b=\(\sqrt{1.6a^2+h^2}-0.675h=\sqrt{1.6×15^2+25^2}\)-0.675×25=14.51 cm

Now,

\(S_c=\frac{0.316P}{h^2}[4 logl/b+1.069]=\frac{0.316×5100}{25^2}\) [4 log 75.86/14.51+1.069]=10.17 kg/cm

^{2}

The warping stress is found using \(St_c=\frac{Eⅇt}{3(1-μ)}\sqrt{\frac{a}{l}}\) .

\(St_c=\frac{2×10^5×10 ×10^{-6}×20}{3(1-0.13)} \sqrt{\frac{15}{75.86}}\)=6.82 kg/cm

^{2}

Therefore, the worst stress combination = S

_{c}+St

_{c}= 10.17 + 6.82 = 16.99 kg/cm

^{2}

14. The combination of stresses would lead to failure when the slab is ______

a) Short

b) Wider

c) Thicker

d) Narrower

View Answer

Explanation: The slab dimensions are in a way responsible for their failure. The combination of stresses can cause the slab to fail when the slab is longer, wider and thinner. This increases the capacity of the slab to carry those stresses and result in cracking, joint widening and even blowup.

15. The rigid pavement is usually designed for edge stress and checked for corner stress.

a) True

b) False

View Answer

Explanation: Considering all the combinations of stress possible, the edge stress is found to be the most critical one. Hence, rigid pavement is designed for edge stress. Corner stress is to be checked for because these tend to curl up during the night and impose stress on the slab.

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