Temperature stresses in the pavement are caused due to the variation

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» B. Cement

Discussion

No Comment Found

Related MCQs

Which temperature stress is generally ignored in the design?
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?
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
How was the temperature stress analysis done by J. Thomlinson?
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.
The temperature differential is affected by the geographical features of the pavement location.
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
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?
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
Frictional stresses are developed due to ______

Temperature stresses in the pavement are caused due to the variation of temperature in ______

Discussion

No Comment Found

Related MCQs

Reply to Comment