Elastic, perfectly plastic bending

It may be seen that with this non-strain-hardening model, the moment still increases beyond the limiting elastic moment and reaches 1.5M e before becoming constant. For this reason, elastic plastic bending is usually a stable process in which the curvature increases uniformly in the sheet without kinking. It may be shown that for materials that do not fit this elastic, perfectly plastic model, for example aged sheet having a stress–strain curve as shown in Figure 1.4, the moment characteristic is different and kinking may occur. In real materials it is very difficult to predict precisely the moment curvature character- istic in the region covered by the bold curve in Figure 6.12 from tensile data. The moment characteristic is extremely sensitive to material properties at very small strain and these properties often are not determined accurately in a tension test. M r 1 Curvature 4 St 2 6 St 2 Figure 6.12 Moment curvature diagram for an elastic, perfectly plastic sheet bent without tension.

6.5.4 Bending of a strain-hardening sheet

If a power law strain-hardening model of the kind shown in Section 6.4.3 and Figure 6.6d is used, the stress distribution will be as shown in Figure 6.13. The whole section is assumed to be deforming plastically and the stress at some distance, y, from the middle surface is σ 1 = K ′ ε n 1 ≈ K ′ y ρ n 6.24 t 2 t 2 y M s 1 Figure 6.13 Stress distribution for a power-law-hardening sheet bent without tension. The equilibrium equation can be written as M = 2K ′ 1 ρ n t 2 y 1 +n dy = K ′ 1 ρ n t n +2 n + 2 2 n +1 6.25 90 Mechanics of Sheet Metal Forming These equations can be combined to give a set of equations for bending a non-linear material, i.e. M I n = σ 1 y n = K ′ 1 ρ n 6.26 where I n = t n +2 2 n +1 n + 2 6.27 The moment, curvature diagram is shown in Figure 6.14. M 1 Moment Curvature r Figure 6.14 Moment curvature diagram for a strain-hardening sheet bent without tension. The power law equation is not a good fit for most materials at very small strains so that Equation 6.26 will not predict the moment curvature characteristic near the elastic, plastic transition Equations 6.26 reduces to some familiar relations for special cases. For the linear elastic model, n = 1, K ′ = E ′ , and we obtain Equations 6.17. For the rigid, perfectly plastic model, n = 0, K ′ = S, and Equation 6.26 reduces to Equation 6.21.

6.5.5 Worked example moments

A hard temper aluminium sheet, 2 mm thick, has a flow stress of 120 MPa, that is approx- imately constant. Determine the moment per unit width to bend the sheet to the limiting elastic state. What is the radius of curvature at this moment? Determine the fully plastic moment if the sheet is bent further. The elastic modulus is 70 GPa and Poisson’s ratio is 0.3. Solution From Equation 6.6, the plane strain plastic bending stress is S = 2 √ 3 σ f = 138.6 MPa Bending of sheet 91