Worked example curving an elastic, perfectly plastic sheet

Moment M M T y × Plastic T y Tension T Elastic r e r a 10 20 30 100 200 300 400 500 Tension, T kNm b Moment, M Nmm 5 10 15 360 370 380 390 400 Tension, T kNm c Moment, M Nmm Figure 10.4 a Effect of tension on the moment on an elastic, perfectly plastic sheet. b Moment versus tension for the worked example in Section 10.2.1. c Moment tension relation for materials with different flow stresses. For the upper curve, the flow stress is 2 greater than that for the lower curve. 140 Mechanics of Sheet Metal Forming The limiting elastic radius of curvature, from Equation 6.19, is ρ e = E ′ t 2S = 220 × 10 9 × 1.5 × 10 −3 2 × 254 × 10 6 = 0.650 m The elastic moment in the sheet as it is bent over the former, from Equation 10.2, is M o = E ′ t 3 12 1 ρ = 220 × 10 9 × 1.5 × 10 −3 3 12 × 2.39 = 25.9 Nm m The fully plastic yielding tension is T y = St = 254 × 10 6 × 1.5 × 10 −3 = 381 kN m The moment in the sheet as the tension is increased, from Equation 10.6, is M = 25.9 1 2 + 3m 4 − m 3 4 Nm m The tension in the sheet is given as a function of m by Equation 10.7, i.e. T = 381 1 − 0.650 4 × 2.39 m + 1 2 Computing the tension and moment in the range −1 ≤ m ≤ 1, one obtains the characteristic shown in Figure 10.4b. We note that the moment remains constant in the elastic range, and then falls rapidly to zero as the tension increases to the yield tension. b Consider the case in which the above process is controlled by setting the tension to the yield tension, 381 kNm. Imagine that new sheet is formed with all conditions the same, except that the yield stress of the sheet is 2 greater; i.e. the yielding tension for the sheet is 389 kNm. If the tension is not changed, there will be some residual moment. Determine approximately the final radius of curvature of the new sheet. Solution. The new plane strain yield stress is S = 1.02 × 254 = 259 MPa and the yield tension is T y = 259 × 10 6 × 1.5 × 10 −3 = 389 kN m The limiting elastic radius of curvature is ρ e = 220 × 10 9 × 1.5 × 10 −3 2 × 259 × 10 6 = 0.637 The tension during stretching is, T = 389 1 − 0.637 4 × 2.39 m + 1 2 kN m In computing a new characteristic, only the tension is changed from the above. The characteristics for the two materials are shown in an enlarged view in Figure 10.4c. Combined bending and tension of sheet 141 For a tension of 381 kNm, the moment in the sheet is approximately 4.8 Nmm. For elastic unloading, the change in curvature is proportional to the change in moment, and from Equation 6.30, 1 ρ = M E ′ I = 4.8 220 × 10 9 1.5 × 10 −3 3 12 = 0.078 m −1 The curvature of the form block is 12.39 = 0.418 m −1 , hence the final curvature is 0.418 − 0.078 = 0.340 m −1 i.e. the final radius of curvature is 10.340 = 2.94, or a change in radius from that of the form block of 2.94-2.392.39100 = 23. This illustrates that the curvature of the sheet is very sensitive to tension if the process is in the elastic, plastic region. For this reason, the sheet is usually overstretched, as mentioned above, to ensure that changes in the strength or thickness of the incoming sheet will not result in springback.

10.3 Bending and stretching a strain-hardening sheet

The previous section assumed that the material did not strain-harden. In practice, sheet having some strain-hardening potential will be used to ensure that it can be stretched beyond the elastic limit in a stable manner. The strain distribution will be similar to that shown in Figure 10.3, but once the whole section has yielded, the stress distribution will not be uniform; the stress will increase from the inner to the outer surface as shown in Figure 10.5. As this stress is not constant, there will be some moment in equilibrium with the section as well as the tension. The stress–strain curve for the material is shown in Figure 10.5a. The range of strain across the section is from just below the mid-surface strain ε 1a , to just above it; the stress–strain relation is assumed to be linear in this range with a slope of |dσdε| ε 1a . The stress and strain distributions are shown in Figure 10.5. The strain is given by ε 1 = ε a + ε b = ε a + y ρ 10.8 The stress is given by σ 1 = σ a + dσ 1 dε 1 ε 1a y ρ 10.9 The moment associated with the stress distribution is M = t 2 −t2 σ a + dσ 1 dε 1 ε 1a · y ρ y dy = dσ 1 dε 1 ρ t 3 12 10.10 On unloading the tension, the sheet will shrink slightly along its middle-surface, but in the two-dimensional, cylindrical case, this will only have a very small effect on the curvature. If the moment unloading process is elastic, Equation 6.30 will give the change in curvature associated with a change in moment of −M. As, for unit width of sheet, I = t 3 12, the proportional change in curvature is 1ρ 1ρ = − |dσ 1 dε 1 | ε 1a E ′ 10.11 142 Mechanics of Sheet Metal Forming a s 1 s a e 1a d s 1 e 1 a d e 1 e 1 2 t 2 t M T y b c d s 1 d e 1 e 1a s a e b = y r e 1 s 1 × e b Figure 10.5 a Stress–strain curve for a strain-hardening material showing the range of stress in a sheet bent and stretched to a mid-surface strain of ε 1a . b The strain distribution when bent over a former of radius ρ , and c the stress distribution. The term dσ 1 dε 1 is the plastic modulus. It is much smaller than the elastic modulus, but in a material obeying a power law strain-hardening model the slope of the curve is dσ 1 dε 1 = n σ 1 ε 1 10.12 At small strains, the slope is greatest and springback could be significant. For this reason, it is preferable to stretch a strain-hardening sheet by a few per cent so that the slope of the stress-strain curve is reduced and springback is less sensitive to strain.

10.3.1 Worked example curving a strain-hardening sheet

A steel sheet has a plane strain stress strain curve fitted by σ 1 = 800 0.0015 + ε 1 0.22 MPa It is stretched over a form block as shown in Figure 10.1. The block is 1.6 m wide and the radius of curvature is 5 m. Determine the springback in terms of change in crown height if the sheet is stretched to a mean plastic strain of a 0.003 0.3, and b 0.015 1.5. E ′ = 220 GPa. Solution. The slope of the stress strain curve is dσ 1 dε 1 = 0.22 × 800 0.0015 + ε 1 0.78 MPa Combined bending and tension of sheet 143