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

Determine the warping stress at the edge of a 20 cm thick pavement if

1.	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
A.	16.6 kg/cm2
B.	15.6 kg/cm2
C.	15.5 kg/cm2
D.	16.5 kg/cm2
Answer» C. 15.5 kg/cm2

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 ______

Discussion

No Comment Found

Related MCQs

Reply to Comment