Question

A sheet-metal part that is 5.0 mm thick, 85 mm long, and 20 mm wide is bent in a wiping die to an included angle of 90o and a bend radius of 7.5 mm. The bend is made in the middle of the 85 mm length, so that the bend axis is 20 mm long. The metal has a yield strength of 220 MPa and a tensile strength of 340 MPa. Compute the force required to bend the part, given that the die opening dimension

Answer 1

To calculate the force, it is necessary to apply the concepts given in the force required to bend, which is determined by the relationship between the constant for wiping die, the tensile strength, the width, the thickness and the die opening dimension as shown in continuation

`F = (k(TS)wt^2)/(D)`

Here,

k = Constant

TS= Tensile strength

w = Width of the part in the bending axis direction

t = Thickness

D = Die opening dimension

k= 0.33 for wiping die, and assigning a value of 8mm for the opening dimension we have,

`F = ((0.33)(340*10^6)(20*10^(-3))(5*10^(-3))^2)/(8*10^(-3))`

`F = 7012.5N`

Therefore the force required to bend the part is 7012.5N