Power law Use of a pre-strain constant

has been hardened in some prior process, this constant indicates a shift in the strain axis corresponding to this amount of strain as shown in Figure 3.4b. In materials which are very nearly fully annealed and for which ε is small, this relation can be obtained by first fitting Equation 3.6 and then, using the same values of Kand n, to determine the value of ε by fitting the curve to the experimentally determined initial yield stress using the equation σ f = Kε n 3.8

3.5.3 Linear strain-hardening

Although the fit is not good, a relation of the form σ = Y + P ε 3.9 may sometimes be used, as shown in Figure 3.4c. This law is appropriate for small ranges of strain.

3.5.4 Constant flow stress rigid, perfectly plastic model

In approximate models, strain-hardening may be neglected and the law σ = Y 3.10 employed. If the strain range of the process is known, the value of Y may be chosen so that the work calculated from the law will equal the work done in the actual process, i.e. the area under the approximate curve will equal that under the real curve and the areas shaded in Figure 3.4d will be equal.

3.5.5 Worked example empirical laws

An aluminium alloy is used in a process in which the effective strain is ε = 0.40. The stress–strain curve is fitted by a law σ = 500ε 0.26 MPa. a Determine the effective stress at the end of the process. b Determine a suitable value for stress in a constant flow stress law. c If the initial yield stress is 100 MPa, determine a suitable value for the constant ε , in the law, σ = Kε + ε n MPa. Compare the stress given by this law and the simple power law at the final strain. Solution a At ε = 0.40, σ = 500 × 0.4 0.26 = 394 MPa. b For equal work done, Y ε = ε Kε n dε = K 1 + n ε 1 +n ε ∴Y = Kε n 1 + n = 394 1.26 = 313 MPa 38 Mechanics of Sheet Metal Forming c By substitution, at ε , σ = 500ε 0.26 = 100 MPa∴ε = 0.20 1 0.26 = 0.002. At ε = 0.40, σ = Kε + ε n gives σ = 395 MPa; i.e. the difference in the laws is neg- ligible at large strains.

3.6 The stress diagram

It has already been mentioned that a diagram in which the strains are plotted, e.g. Figure 3.2c, is valuable in the study of a process. In the same way, a diagram in which the stress state associated with each strain point is shown is very useful in understanding the forces involved in a process. Such a diagram is shown in Figure 3.5. Like Figure 3.3a, this is not a diagram for a particular process, but is used to illustrate the link between the strain and stress diagrams. Also, contours of equal effective stress are shown, which are of course yield loci for particular values of flow stress. During deformation, plastic flow will start from the initial yield locus shown as a continuous line, i.e. when σ = σ f and the loading path will be along a radial line of slope 1α. B A C D E a = − ∞ a = 12 a = − 1 a = 1 a = s 1 s 2 s f 1 s f a Figure 3.5 The processes shown in the strain space, Figure 3.2, illustrated here in the stress space. The current yield ellipse is shown as a broken line.. To plot a point in this diagram, the stress ratio is calculated from the strain ratio, Equation 2.14. The effective strain is determined from Equation 2.19c, and from the known material law, the effective stress determined and the principal stresses calculated from Equation 2.18b. The current state of stress is shown as a point on the ellipse given as a broken line. This yield locus intercepts the axes at ±σ f . The principal stresses are σ 1 ; σ 2 = α.σ 1 and σ 3 = 0 3.11 and each path in the strain diagram, Figure 3.3a has a corresponding path in the stress diagram as detailed below. Deformation of sheet in plane stress 39