Pavement Design Questions and Answers – Combined Stresses in Rigid Pavement

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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

Answer: a
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

Answer: b
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

Answer: b
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.
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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

Answer: c
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

Answer: a
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

Answer: c
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

Answer: a
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.
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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/cm2
Warping stress = 8.56 kg/cm2
Frictional stress = 0.77 kg/cm2
a) 39.06 kg/cm2
b) 39.83 kg/cm2
c) 31.27 kg/cm2
d) 38.29 kg/cm2
View Answer

Answer: a
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/cm2.

9. A cc slab of thickness 20 cm is constructed over a base of granular material with a modulus of reaction 10 kg/cm3. 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 3×105 kg/cm2, 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/cm2
b) 59.51 kg/cm2
c) 59.15 kg/cm2
d) 51.95 kg/cm2
View Answer

Answer: d
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 log⁡l/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,
\(S_e=\frac{0.572P}{h^2}[4 log⁡l/b+0.359] =\frac{0.572×5100}{20^2}\) [4 log⁡67.26/14.07+0.359]=22.44 kg/cm2
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 Cx and Cy are called Bradbury’s coefficients and their values can be obtained using values given in IRC 58:2002. To find Cx and Cy, the ratios \(\frac{L_x}{l}\) for Cx and \(\frac{L_y}{l}\) for Cy have to be found out. Lx is the spacing between the transverse joints = 800 cm and Ly 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
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Therefore, for \(\frac{L_x}{l}\)= 11.89, the value would be \(\frac{C_x-1.05}{11.89-11}=\frac{1-1.05}{12-11}\), Cx=1.01 and for \(\frac{L_y}{l}\)= 5.95, the value would be \(\frac{C_y-0.72}{5.95-5}=\frac{0.92-0.72}{6-5}\), Cy=0.91.
Cx 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/cm2
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/m3.
\(S_f=\frac{2400×4×1.5}{2×10^4}\)=0.72 kg/cm2
Therefore, the worst stress combination would be Se+Ste+Sf = 22.44 + 28.79 + 0.72 = 51.95 kg/cm2

10. The load stress is the least in which region?
a) Corner
b) Edge
c) Interior
d) Bottom
View Answer

Answer: c
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/cm2
Warping stress = 30.31 kg/cm2
Frictional stress = 0.89 kg/cm2
a) 65.09 kg/cm2
b) 65.90 kg/cm2
c) 69.50 kg/cm2
d) 69.05 kg/cm2
View Answer

Answer: a
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/cm2.

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

Answer: b
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/cm3. 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 2×105 kg/cm2, 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/cm2
b) 15.99 kg/cm2
c) 16.99 kg/cm2
d) 15 kg/cm2
View Answer

Answer: c
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 log⁡l/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,
\(S_c=\frac{0.316P}{h^2}[4 log⁡l/b+1.069]=\frac{0.316×5100}{25^2}\) [4 log⁡ 75.86/14.51+1.069]=10.17 kg/cm2
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/cm2
Therefore, the worst stress combination = Sc+Stc = 10.17 + 6.82 = 16.99 kg/cm2

14. The combination of stresses would lead to failure when the slab is ______
a) Short
b) Wider
c) Thicker
d) Narrower
View Answer

Answer: b
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

Answer: a
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.

Sanfoundry Global Education & Learning Series – Pavement Design.

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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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