# Pavement Design Questions and Answers – Temperature Stresses in Rigid Pavement

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

1. Temperature stresses in the pavement are caused due to the variation of temperature in ______
a) Slab
b) Cement
d) Sub-base

Explanation: In the rigid pavement, the topmost layer is the cement concrete slab. So, when the slab is subjected to changes in temperature, it induced the stresses due to change in temperature in the pavement.

2. The seasonal variation of temperature causes the variation in temperature across the depth of the slab.
a) True
b) False

Explanation: The concept of temperature change can be attributed to its effect due to daily and seasonal variations. The temperature change due to daily variations causes the variation in temperature across the depth of the slab. Overall temperature change is caused by seasonal variations.

3. 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

Explanation: The change in temperature causes two types of stresses in the rigid pavement. They are warping stress and frictional stress. Both are computed using specific equations and their combinations can also be assessed.
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4. 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

Explanation: Whenever the top and bottom layers of the slab possess different temperature at the same time, then the warping or the curling of the slab either downwards or upwards takes place. This condition causes the development of warping stresses. There is no specific limit for the temperature differential.

5. 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) $$S_f=\frac{wLf}{2×10^5}$$

Explanation: $$S_f=\frac{wLf}{2×10^4}$$ is the equation to find the frictional stress in the rigid pavement. In the equation, Sf is the frictional stress, W is the unit weight of concrete, L is the length of the slab and f is the coefficient of friction. The factor 2×104 is used for unit conversions.

6. 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

Explanation: Frictional stress is caused due to the seasonal variation in temperature. As temperature changes in the summer and winter season, the slab expands or contracts. This leads to the development of frictional stress to resist the movement of the slab. The stress is developed opposite to the movement of the slab.

7. Determine the warping stress at the edge of a 20 cm thick pavement if the following data is provided.
Spacing of transverse joint = 10 m
Spacing of longitudinal joint = 3.6 m
Modulus of subgrade reaction = 6 kg/cm3
Temperature differential = 0.5°C per cm of slab thickness
Radius of contact area = 15 cm
Poisson’s ratio = 0.15
Modulus of elasticity = 3×cm5 kg/cm2
Thermal 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

Explanation: There are two equations to compute the warping stress at the edge of the slab. The equation giving the highest stress value is taken as the answer. The equations are $$St_e=\frac{C_x Eet}{2}$$ and $$St_e=\frac{C_y Eet}{2}$$. The terms Cx and Cy are called Bradbury’s coefficients in X and Y direction respectively. These are obtained using graphs if the length of joints and radius of relative stiffness are known. E is the modulus of elasticity; e is the thermal coefficient and t is the temperature differential.
First, the parameter $$l=\left[\frac{Eh^3}{12k(1-μ^2)}\right]^{1∕4}$$ is found out.
$$l=\left[\frac{3×10^5×20^3}{12×6×(1-0.15^2)}\right]^{1∕4}$$=76.42 cm
Now 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 transverse joints = 1000 cm and Ly is the spacing between longitudinal joints = 360 cm.
$$\frac{L_x}{l}=\frac{1000}{76.42}=13.09$$ and $$\frac{L_y}{l}=\frac{360}{76.42}=42.71$$
The design graphs have been given by Bradbury and sample of the same is as below

From the graph, the value of Cx=1.04 and Cy=0.65. Now, Cx has a higher value, therefore, the equation giving the highest stress value would be $$St_e=\frac{C_x Eet}{2}$$
$$St_e=\frac{C_x Eet}{2}=\frac{1.04×3×10^5×10×10^{-6}×0.5×20}{2}$$=15.6 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

Explanation: The equation to find the frictional stress is $$S_f=\frac{WLf}{2×10^4}$$. The length of the slab is the same as the spacing between the contraction joints.
$$S_f=\frac{2400×4.2×1.3}{2×10^4}$$ =0.66 kg/cm2

9. Find the warping stress at the corner of the pavement for the data given below.
Pavement thickness = 25 cm
Radius of relative stiffness = 64.50 cm
Temperature differential = 0.6°C per cm of slab thickness
Radius of contact area = 15 cm
Poisson’s ratio = 0.15
Modulus of elasticity = 2.1×105 kg/cm2
Thermal 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

Explanation: The equation for the computation of corner stress is $$St_c=\frac{Eⅇt}{3(1-μ)}\sqrt{a}{l}$$ . All the terms required are given in the question as:
E = 2.1×105 kg/cm2, e = 10×10-6 /°C, t = 0.6°C X 25 cm = 15°C, a = 15 cm, l = 64.50 cm and μ = 0.15
$$St_c=\frac{2.1×10^5×10 ×10^{-6}×15}{3(1-0.15)}\sqrt{15}{64.5}$$=5.96 kg/cm2

10. The temperature differential is affected by the geographical features of the pavement location.
a) True
b) False

Explanation: The temperature differential is a function of the solar radiation received by the pavement, the losses due to wind velocity and thermal diffusivity of concrete. These are all geographical features of the location of the pavement.

11. Find the warping stress in the interior of the pavement for the data given below.
Temperature differential, t = 17°C
Poisson’s ratio, μ = 0.15
Modulus of elasticity, E = 2×105 kg/cm2
Thermal coefficient, e = 10×10-6 /°C
Cx=0.75 and Cy=0.56.
a) 15.5 kg/cm2
b) 14 kg/cm2
c) 15 kg/cm2
d) 14.5 kg/cm2

Explanation: $$St_i=\frac{Eⅇt}{2}[\frac{C_x+μC_y}{(1-μ^2)}]$$ is the formula to find the interior stress. After substituting all the terms in the question in the formula,
$$St_i=\frac{2×10^5×10×10^{-6}×17}{2}[\frac{0.75+0.15×0.56}{(1-0.15^2)}]$$=14.5 kg/cm2.

12. How was the temperature stress analysis done by J. Thomlinson?
a) Thermometers
b) Thermocouples
c) Thermostats
d) Thermocline

Explanation: J. Thomlinson’s temperature stress analysis was an analytical method formulated in 1940. The analysis involved the use of thermocouple for the actual measurement of the pavement layer temperature. The analysis yields result closer to the experimental data.

13. Determine the Bradbury coefficients of a 25 cm thick pavement if the following data is provided.
Spacing of transverse joint = 10 m
Spacing of longitudinal joint = 4 m
Modulus of subgrade reaction = 7 kg/cm3
Radius of contact area = 15 cm
Poisson’s ratio = 0.15
Modulus 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

Explanation: To find the Bradbury’s coefficients, Cx and Cy from the design graph, we need to compute the radius of relative stiffness.
$$l=[\frac{Eh^3}{12k(1-μ^2)}]^{1⁄4}=[\frac{3×10^5×25^3}{12×7×(1-0.15^2)}]^{1⁄4}$$=86.92 cm
The ratios are found out as below:
$$\frac{L_x}{l}=\frac{1000}{86.92}$$=11.50 and $$\frac{L_y}{l}=\frac{400}{86.92}$$=4.60
Using the sample graph, the Bradbury’s coefficients Cx and Cy are found out as below:

Therefore, Cx=1.05 and Cy=0.64.

14. 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

Explanation: The equation to find the frictional stress is $$S_f=\frac{WLf}{2×10^4}$$. The length of the slab is the same as the spacing between the contraction joints.
$$S_f=\frac{2400×5×1.5}{2×10^4}$$=0.9 kg/cm2

15. Which temperature stress is generally ignored in the design?
a) Frictional stress
b) Interior stress
c) Edge stress
d) Corner stress

Explanation: As mentioned in IRC 58:2002, the corners of a cement concrete slab are relatively free to warp and so the temperature stress in that region is negligible. Hence, temperature stress is generally ignored.

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