Question

A stretch of 1 km new concrete pavement is being laid in a 100 mm thick layer. The ambient temperature when laying the concrete is 20°C, and you can assume that the concrete does not have any internal stress at this temperature. The thermal expansion coefficient of concrete is 10μm/m°C, and its elastic modulus is 20 GPa. Suppose all deformation of the concrete is restrained (the concrete cannot move), and suppose that the concrete behaves as a linear elastic material. Knowing its tensile strength is 10 MPa, at what temperature does this concrete crack?

Answer 1

Answer:

70 °C

Explanation:

Note : 1 GPA = 1000MPa ; 1 μ = 10⁻⁶

Strain (∈) = δl / l = ∝ ΔT

Elastic modulus = Direct stress (strength)/Strain

         20 * 10³  = 10 / ∝ ΔT

ΔT  =  10 /  20 * 10³ ∝

     = 10 / ( 20 * 10³ *10⁻⁶)

      =  10/0.02 = 50

      = 50 °C

Therefore Temperature for concrete crack = ambient temperature + ΔT

20°C + 50 °C

                                                                        = 70 °C