Bending of a strain-hardening sheet

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 From Equation 6.10, the elastic modulus in plane strain is E ′ = E 1 − υ 2 = 76.9 GPa From Equation 6.18, the limiting elastic bending moment is M e = St 2 6 = 138.6 × 10 6 2 × 10 −3 2 6 = 92.3 Nmm The radius of curvature, from Equation 6.19, is ρ e = E ′ t 2S = 76.9 × 10 9 × 2 × 10 −3 2 × 138.6 × 10 6 = 0.555 m or 555 mm The fully plastic moment from Equation 6.21 is M p = St 2 4 = 3 2 M e = 138 Nmm

6.6 Elastic unloading and springback

If a sheet is bent by a moment to a particular curvature, as shown in Figure 6.15, and the moment then released, there will be a change in curvature and bend angle. The length of the mid-surface is l = ρθ This will remain unchanged during unloading as the stress and strain at the middle surface are zero. From this, we obtain θ = l 1 ρ 6.28 Differentiating Equation 6.28, in which l = constant, we obtain θ θ = 1ρ 1ρ 6.29 M M q q + ∆q r + ∆r r Figure 6.15 Unloading a sheet that has been bent by a moment without tension. 92 Mechanics of Sheet Metal Forming