Strain distribution Small radius bends

not the case. We consider a fibre y from the middle surface in the undeformed state as shown in Figure 6.20. Equating the shaded volumes shown, θ 2 r 2 − ρ − t 2 2 = y + t 2 l 6.39 Given that, θ = lρ = l ρ , the radius of the fibre initially at y is r = 2ρy + ρ 2 + t 2 4 6.40 The length of the fibre, A ′ B ′ , is l = rθ, and substituting θ = l ρ we obtain l l = 1 + 2y ρ + t 2ρ 2 6.41 The change in fibre length on bending is shown in Figure 6.21a. Fibres initially above the middle surface will always increase in length. Fibres below the middle surface will decrease in length initially, but may then increase. The minimum length is denoted by the point B in Figures 6.21a and b. The minimum length is found by differentiating Equation 6.41 with respect to curvature 1ρ and equating to zero. From this, the strain in a fibre begins to reverse when t ρ = − 4y t 6.42 Substituting in Equation 6.41, the minimum length of such a fibre is l l = 1 − 2y t 2 6.43 and the radius of the fibre at B that has reached its minimum length is r b = ρ 2 − t 2 2 = RR 6.44 The importance of this strain reversal is that it must be taken into account in determining the effective strain in a material element at a particular distance from the middle surface. If there is no reversal, the effective strain can be determined approximately from the initial and final lengths of the fibre. If there is a reversal, the effective strain integral should be integrated along the whole deformation path. In Figure 6.21c, the bold curve on the right shows the effective strain determined from the above analysis. The broken line is that determined from the simple, large bend ratio analysis. There is a significant difference due to two factors: • appropriate integration of the effective strain; and • the non-linear distribution of strain derived from Equation 6.41. The curves in Figure 6.21c are calculated for a bend ratio ρt = 23. At the inner surface, y = −t2, the strain calculated from Equation 6.41 is ε 1 = −1.4. In the simple analysis ε 1 = y ρ = −t 2ρ = −0.75 Bending of sheet 97 2.0 0.5 0.25 0.25 0.50 0.75 −0.15 −0.25 −0.4 −0.5 1.5 0.5 0.5 O C B 0.25 0.50 0.75 −1 1 ∋ ∋ ∋ −0.5 A l l min yt l l 1 = ln ll t 2r −0.15 −0.25 −0.40 t 2r y t = y t = 0.25 A B C Approximate analysis Strain reversal a b c Figure 6.21 a Changes in length of fibres at different initial distances y t from the middle surface. b Changes in distance from the middle surface during bending. c Axial strain ε 1 and effective strain ε in sheet bent to a bend ratio of ρt = 23; the effective strain derived from the simple, large bend ratio analysis is shown by the broken line.

6.7.2 Stress distribution in small radius bends

In the simple analysis, the stresses assumed in the bend region are those given by Equation 6.5. In small radius bends, the through-thickness stress cannot be assumed to be zero. As plane strain is assumed, the strain and stress state, as shown in Figure 6.22, is ε 1 ; ε 2 = 0; ε 3 = −ε 1 σ 1 ; σ 2 = σ 1 + σ 3 2 ; σ 3 6.45 The equilibrium equation for forces in the through-thickness radial direction is σ 3 + dσ 3 r + dr dθ1 − σ 3 r dθ + σ 1 dr dθ 1 = 0 i.e. dσ 3 dr − σ 1 − σ 3 r = 0 6.46 98 Mechanics of Sheet Metal Forming R R dq d r s 3 s 1 s 3 + ds 3 r Figure 6.22 Stresses on an element at a radius r in a small radius bend. To investigate the stresses, we assume a Tresca yield criterion as in Section 2.4.4. In the region where the bending stress is tensile, the greatest stress is σ 1 and the least is σ 3 ; therefore, σ 1 − σ 3 = S = σ f and Equation 6.46 becomes dσ 3 dr − S r = 0 If the stress on the outer surface is zero, integrating between the limits r = R and r, we obtain σ 3 = −S ln R r 6.47a In the region where the bending stress is compressive, the yield criterion gives that σ 1 − σ 3 = −S = −σ f Integrating Equation 6.46 between the limits r = R and r, and assuming that the stress on the inner surface is zero, we obtain σ 3 = −S ln r R 6.47b The stress distributions are shown schematically in Figure 6.23. It should be noted that these are for a non-strain-hardening material in which S is constant. The through thickness stresses given by Equations 6.67a and b are equal at a radius of r b = RR o As noted earlier, this is the radius at which the bending strain reverses. Bending of sheet 99