Question

A polystyrene component must not fail when a tensile stress of 1.25 MPa (180 psi) is applied. Determine the maximum allowable surface crack length if the surface energy of polystyrene is 0.50 J/m2 (2.86 103 in.-lbf/in.2 ). Assume a modulus of elasticity of 3.0 GPa (0.435 106 psi).

Answer 1

Answer:

Maximum allowable surface crack length =  l =  0.000611m

Explanation:

Tensile stress = σ = 1.25 MPa = 1.25 × 10^6 N/m2

Surface energy of polystyrene = Es = 0.50 J/m2

Elastic Mudolus = E = 3 GPa = 3 × 10^9 N/m2

Maximum allowable surface crack length = l = ?

We know that:

       l = 2EEs/πσ^2 ....... (i)

By putttin the values in equation (i)

       l = (2)(0.50)(3 × 10^9)/(3.14)(1.25 × 10^6)^2

       l =  0.000611m

Hence, the Maximum allowable surface crack length will be 0.000611m.

Answer 2

Explanation:

Below is an attachment containing the solution.