Residual stresses after unloading

≡ S −3 S 2 M p M = 0 − M p − S 2 + Figure 6.18 Residual stress distribution after unloading from a fully plastic moment. side of the bend would have a significant compressive residual stress at the surface and there would be a residual tensile stress on the inner surface.

6.6.3 Reverse bending

If a sheet has been bent to a fully plastic state and unloaded, it is interesting to see what reverse bending is required to cause renewed plastic deformation. From Figure 6.16, it may be seen that the change in stress required at the outer fibre to just start yielding is −2S. Substituting in Equation 6.30, shows that the change in moment is M = − 2S t 2 t 3 12 = − St 2 3 6.35 The moment for reverse yielding is therefore M rev. = St 2 1 4 − 1 3 = − St 2 12 = − M e 2 6.36 Thus yielding starts at only half the initial yield moment as shown in Figure 6.19. This softening effect is important as there are a number of processes in which sheet goes through bend–unbend and reverse bend cycles. The actual softening is likely to be greater than that calculated above as most materials will also have some Bauschinger effect and yield at a reverse stress of magnitude less than S. M Moment Curvature M e = 4 M p = 1r 2 12 St 2 − − M p St 2 6 St 2 M e = − Figure 6.19 Reverse bending of an elastic, perfectly plastic sheet. Bending of sheet 95

6.7 Small radius bends

6.7.1 Strain distribution

In the previous sections, the strain in bending was assumed to be a linear function of the distance from the middle surface. If the radius of the bend is approximately the same as the sheet thickness, a more refined analysis is necessary. In Figure 6.20, a length l of sheet of thickness t is bent under plane strain and constant thickness conditions to a middle surface radius of ρ. In the deformed shape, the length of the middle surface is l a = ρθ A l B C D y t 2 y A′ B′ C′ D′ q R r r b r t Figure 6.20 Element of sheet bent to a small radius bend. and the volume of the deformed element is, θ 2 R 2 − R 2 1 where R = ρ + t2 is the radius of the outer surface and R = ρ − t2 is the radius of the inner surface. As the volume of the element remains constant, we obtain from the above l t 1 = θ 2 R 2 − R 2 1 = θρt i.e. l = θρ 6.37 If the length of the middle surface in the deformed condition is l a then, l a = ρθ = l 6.38 i.e. the length of the middle surface does not change. In the previous simple analysis, it was assumed that a fibre at some distance y from the middle surface remained at that distance during bending. In small radius bends, this is 96 Mechanics of Sheet Metal Forming