The Levy–Mises flow rule

that it is the deviatoric components that will be the ones associated with the shape change. This is the hypothesis of the Levy–Mises Flow Rule. This states that the ratio of the strain increments will be the same as the ratio of the deviatoric stresses, i.e. dε 1 σ ′ 1 = dε 2 σ ′ 2 = dε 3 σ ′ 3 2.13a or dε 1 2 − α = dε 2 2α − 1 = dε 3 −1 + α 2.13b If a material element is deforming in a plane stress, proportional process, as described by Equation 2.6, then Equation 2.13b can be integrated and expressed in terms of the natural or true strains, i.e. ε 1 2 − α = ε 2 2α − 1 = βε 1 2α − 1 = ε 3 −1 + α = −1 + βε 1 −1 + α 2.13c

2.5.2 Relation between the stress and strain ratios

From the above, we obtain the relation between the stress and strain ratios: α = 2β + 1 2 + β and β = 2α − 1 2 − α 2.14 It may be seen that while the flow rule gives the relation between the stress and strain ratios, it does not indicate the magnitude of the strains. If the element deforms under a given stress state i.e. α is known the ratio of the strains can be found from Equation 2.13, or 2.14. The relationship can be illustrated for different load paths as shown in Figure 2.9; the small arrows show the ratio of the principal strain increments and the lines radiating from the origin indicate the loading path on an element. It may be seen that each of these strain increment vectors is perpendicular to the von Mises yield locus. It is possible to predict this from considerations of energy or work. 1 12 1 1 −1 1 de 2 = −2de 1 de 2 = −de 1 de 2 = 0 s 1 , de 1 s 2 , de 2 de 2 = −de 1 2 de 2 = de 1 −s f Figure 2.9 Diagram showing the strain increment components for different stress states around the von Mises yield locus. Sheet deformation processes 23

2.5.3 Worked example stress state

The current flow stress of a material element is 300 MPa. In a deformation process, the principal strain increments are 0.012 and 0.007 in the 1 and 2 directions respectively. Determine the principal stresses associated with this in a plane stress process. Solution dε 2 = βdε 1 , ∴β = dε 2 dε 1 = 0.0070.012 = 0.583. ∴α = 2β + 12 + β = 0.839 ∴σ 1 = σ f 1 − α + α 2 = 323 MPa; σ 2 = α.σ 1 = 271 MPa

2.6 Work of plastic deformation

If we consider a unit principal element as shown in Figure 2.10, then for a small defor- mation, each side of the unit cube will move by an amount, 1 × dε 1 ; 1 × dε 2 ; etc. 1 1 × 1 × s 1 1 × 1 × s 2 1 × de 1 1 × 1 × s 2 Figure 2.10 Diagram of a principal element of unit side, showing the force acting on a face and its displacement during a small deformation. and as the force on each face is σ 1 × 1 × 1, etc., the work done in deforming the unit element is dW vol. = σ 1 dε 1 + σ 2 dε 2 + σ 3 dε 3 2.15a For a plane stress process, this becomes W vol. = ε 1 o σ 1 dε 1 + ε 2 o σ 2 dε 2 2.15b 24 Mechanics of Sheet Metal Forming