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 90 degrees and a bend radius of 7.5 mm. The bend is 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 requires to bend the part, given the die opening of 8 mm.

Answer 1

Answer:

force required to bend the part is 7012.5 N

Explanation:

force required to bent  can be calculated by using following relation$F = \frac{k(TS)wt^2}{D}$

Where,

k is constant = 0.33 for wiping die

TS = tensile strength = 340\times 10^6

w = width of part = 20 mm = 0.020 m

t =  thickness  = 5.00 mm = 0.005 m

D = opening dimension of die =  8mm = 0.008 m

putting all value in the above formula

$F = \frac{0.33\times 340\times 10^6\times 20\times 10^{-3}\times (5 \times 10^{-3})^2}{8\times 10^{-3}}$

F = 7012.5 N

Therefore force required to bend the part is 7012.5 N

Answer 2

Answer:

F=7.012 KN

Explanation:

We know that force require for bending is

$F=K\dfrac{\sigma_twt^2}{D}$

Where

K=Constant

t=Thickness

D=Die opening

w=Width

$\sigma_t$=Tensile strength

Here K= 0.33

Now by putting the all given values

$F=K\dfrac{\sigma_twt^2}{D}$

$F=0.33\times \dfrac{340\times 10^6\times 0.02\times (0.005)^2}{0.008}$

F=7012.5 N

F=7.012 KN