PSI - Issue 21

Tuncay Yalçinkaya et al. / Procedia Structural Integrity 21 (2019) 52–60 Yalc¸inkaya et al. / Structural Integrity Procedia 00 (2019) 000–000

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relations ˙ ε z = ˙ h / h , ˙ ε θ = ˙ u / r , ˙ ε r = ˙ u / dr with the boundary condition ( ˙ u = 0 at r = l ), we get, ˙ u = ( ˙ ε z r / 2)( l 2 / r 2 − 1) ˙ ε θ = ˙ u r = ˙ ε z 2 ( l 2 r 2 − 1 ) and ˙ ε r = − ( ˙ ε θ + ˙ ε z ) = − ˙ ε z 2 ( l 2 r 2 + 1 ) Then, using (1) e ff ective strain rate can be written as, ˙ ε e = √ 2 9 [ ( ˙ ε r − ˙ ε θ ) 2 + ( ˙ ε r − ˙ ε z ) (2) Apply the upper bound theorem for a perfectly plastic material which states: work done by the limit load is smaller or equal to the integral energy dissipation of the e ff ective strain rate at yield stress (11). π l 2 T n ˙ δ n ≤ ∫ l a ˙ ε e σ y 2 π rhdr (3) Noting ˙ δ n = ˙ h and ˙ ε z = ˙ h h , and substituting (1) into (11), we get, T n ≤ σ y ∫ 1 f √ 1 + ( 1 3 ν 2 ) d ν where ν = r 2 / l 2 and d ν = 2 r / l 2 dr (4) with f = a 2 / l 2 being the area fraction of pores. Now, we have two ways of evaluating the integral in (see e.g. 4)(Hardy et al. (1978)). Minkowski inequality: ( ∫ ( f + g ) k dx ) 1 k > ( ∫ f k dx ) 1 k + ( ∫ g k dx ) 1 k or Jensen’s inequality: ∫ ( f + g ) k dx < ∫ f k dx + ∫ g k dx Using the Jensen’s inequality retains the formal nature of the upper bound theorem, while the Minkowski inequality is a better approximation of the integral but is the opposite sense of the upper bound theorem. Additionally, Minkowski inequality results in a continuous function for mixed-mode loading while Jensen’s inequality results in a discontinuos one. In short, depending on the inequality we get di ff erent traction-separation equations each having their merits and demerits which can be calibrated with experiments. Both ways will be shown in this paper. Applying Jensen’s inequality to (4) gives, T n ≤ σ y ( (1 − f ) + ( 1 √ 3 ln 1 f ) ) (5) or the yield function, g = T n (1 − f ) + 1 √ 3 ln 1 f − σ y = σ − σ y (6) Applying Minkowski inequality to (4) gives, T n ≈ σ y ( (1 − f ) 2 + ( 1 √ 3 ln 1 f ) 2 ) 1 2 (7) 2 + ( ˙ ε θ − ˙ ε z ) 2 ] = ˙ ε z √ 1 + l 4 3 r 4 (1)

which is no longer an upper bound, but an approximation. Or the yield function, g = [ T 2 n (1 − f ) 2 + ( 1 √ 3 ln 1 f ) 2 ] 1 2 − σ y = σ − σ y

(8)

Assuming the matrix is incompressible, growth of RVE results in the growth of pore, meaning d f / (1 − f ) = dh / h , and assume f = f 0 when h = h 0 . Taking the integral and substituting to δ n = h − h 0 gives f = ( δ n + h 0 f 0 ) / ( δ n + h 0 ). Then, substituting f to (6) gives, T n = σ y [ h 0 (1 − f 0 ) ( δ n + h 0 ) + 1 √ 3 ln ( ( δ n + h 0 ) ( δ n + h 0 f 0 ) ) ] (9)