Benchmark 1 - Failure Prediction after Cup Drawing, Reverse Redrawing and Expansion Part A: Benchmark Description

Abstract

Abstract. This Benchmark is designed to predict the fracture of a food can after drawing, reverse redrawing and expansion. The aim is to assess different sheet metal forming difficulties such as plastic anisotropic earing and failure models (strain and stress based Forming Limit Diagrams) under complex nonlinear strain paths. To study these effects, two distinct materials, TH330 steel (unstoved) and AA5352 aluminum alloy are considered in this Benchmark. Problem description, material properties, and simulation reports with experimental data are summarized.

Material characterization and damage assessment of an AA5352 aluminium alloy using digital image correlation

“…Besides, Table 5 presents the mean value of calculated outcome compared with the ones reported by the literature. 4,35 In consequence, the amounts of n and r 90 acquired from DIC analysis agreed well with the literature. Small deviations verified the results and therefore they could be considered as a robust experimental data adopted for the future analyses of the AA5352 alloys.…”

Influence of the Replacement of the Actual Plastic Orthotropy with Various Approximations of Normal Anisotropy on Residual Stresses and Strains in a Thin Disk Subjected to External Pressure

“…The numerical example provided in this section evaluates the effect of replacing Hill's quadratic yield criterion with several transversely isotropic yield criteria for a disk of inner radius a = 0.35 made of aluminum alloy 5352. This alloy's mechanical properties involved in Yield Criterion (3) are available in the literature [34]. One of the principal axes of anisotropy coincides with the axial direction independently of the orientation of the two other axes.…”

“…One of the principal axes of anisotropy coincides with the axial direction independently of the orientation of the two other axes. The left-hand side of Yield Criterion (3) is known from [34]. Then, Z = 220 MPa and the yield stresses in the directions of the other two principal axes of anisotropy are 198 and 279 MPa.…”

“…Having found the dependence of ε p θ on ψ, one can determine the other plastic strains from Equation (34) and the total strains from Equation (38) as functions of ψ. These functions and the third equation in (15) combine to supply the dependence of all these strains on ρ in parametric form.…”

Effect of the anisotropic yield condition on the predicted distribution of residual stresses in a thin disk

“…Величина s 0 в (19) и (20), включая выражение величины p через P, определяется в зависимости от используемого условия текучести. Решение при упругой разгрузке имеет силу, если выполняются неравенства осреднения условия текучести на распределение остаточных напряжений рассмотрим диск внутреннего радиуса a = 0 35 , , который изготовлен из алюминиевого сплава 5352, обладающего следующими свойствами X = = 197,59 МПа, Y = 278 86 , МПа, R 0 0 535 = , , R 15 = = 0,465, R 30 0 655 = , , R 45 1 105 = , , R 60 1 415 = , , R 75 1 595 = , , R 90 2 270= ,[15], и нагружен давлением P = 140 МПа. Расчеты проведены для следующих случаев: предел текучести в направлении радиальной координаты равен X ( ) ;…”