Pavement Design Multiple Choice Questions 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
c) Subgrade
d) Sub-base
Answer: a
Clarification: 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
Answer: b
Clarification: 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
Answer: c
Clarification: 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.
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
Answer: a
Clarification: 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})
Answer: c
Clarification: (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
Answer: d
Clarification: 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
Answer: b
Clarification: 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