Bending a sheet in a vee-die

F R A t P P a y a−y a B B B MP 1r O r A P F1 B Punch Die Resultant of F 1 and P H 1 Moment arm of B Moment arm of B′ B′ R a b Figure 6.29 Bending a strip in a vee-die with a punch of nose radius R. a At the start of the process. b When the punch has nearly reached the bottom of its stroke. Bending of sheet 105 In the initial stages, the curvature of the sheet at the nose of the punch will be less than the nose radius, as shown in Figure 6.29a. The curvature is given by the point B in the bending line construction shown. As the bending progresses, the punch force will increase and the curvature at the point of contact increase until it just matches the punch curvature. On further bending, the point of contact with the punch will move away from the nose to some point B as shown in Figure 6.29 b. Here we consider only a frictionless condition, so that the force is normal to the tool at the point of contact. It is seen that there is a difference between the line of action of the force exerted by the die at point A and that through the point of contact B with the punch. These forces converge as shown, and by symmetry, their resultant must be horizontal, i.e. the force H . As the moment arm of the force bending the sheet at the centre-line B ′ is greater than that at the punch contact B, the curvature at B ′ must be greater than at B, and there will be a gap between the sheet and the punch at B ′ . If close conformity between the punch and the sheet is required, the vee-die is made with a radius at the bottom to match the punch and a large force is applied at the end of the process. A problem with such an arrangement is that small variations in thickness or strength in the sheet or in friction may cause appreciable changes in springback.

6.10 Exercises

Ex. 6.1 A steel strip, 50 mm wide and 2 mm thick is bent by a pure moment. The plane strain–stress curve follows the law σ = 200ε GPa in the elastic range, and σ = 250 MPa constant in the plastic range. a Determine the limiting elastic curvature. b Construct a moment curvature diagram for the strip in the range 0 1 ρ 8 m −1 c If the strip is wound without tension onto a former of 800 mm diameter, determine the radius of curvature after release: • using the approximate relation 1 ρ = −3 σ f Et ; • using the diagram in b. Ans : a1 .25 m −1 , c1 .6 m vs.1 .28 m. Ex. 6.2. A strip of length l and thickness t is bent to a complete circle. Under load the circumference is l. The material is elastic, perfectly plastic with a constant plane strain yield stress S and elastic modulus E ′ . Calculate the gap between the ends of the strip after unloading. Assume that the strip is fully plastic when bent and that the final gap is small. Ans : 3 2 π l 2 t S E ′ Ex. 6.3 A strip of sheet steel, 2 mm thick and 200 mm wide, is to be bent in a die under conditions of zero friction and zero axial tension. The radius of curvature of the die is 80 mm. The plastic properties of the material are σ = 600ε 0.22 MPa. The elastic properties are E = 200 GPa and Poissons ratio = 0.3. a Find the radius after springback. [Ans : 92 mm] 106 Mechanics of Sheet Metal Forming Ex. 6.4 Springback in bending is given by Equation 6.32. If Aluminum is substituted for steel in identical applications with the same bend radius and sheet thickness, compare the difference in springback for bend ratio of a 10 and b 5 respectively. Al: K = 205 MPa, n = 0.2, density: ρ = 2.7 × 10 3 kgm 3 ; E = 75 GPa, ν = 0.3. Steel: K = 530 MPa, n = 0.26, density: ρ = 7.9 × 10 3 kgm 3 ; E = 190 GPa, ν = 0.3. Ans : Springback ratio, Al to Steel:a1 .173, b1 .125 . Bending of sheet 107