Respuesta :
Answer:
force required to bend the part is 7012.5 N
Explanation:
force required to bent can be calculated by using following relation[tex]F = \frac{k(TS)wt^2}{D}[/tex]
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
[tex]F = \frac{0.33\times 340\times 10^6\times 20\times 10^{-3}\times (5 \times 10^{-3})^2}{8\times 10^{-3}}[/tex]
F = 7012.5 N
Therefore force required to bend the part is 7012.5 N
Answer:
F=7.012 KN
Explanation:
We know that force require for bending is
[tex]F=K\dfrac{\sigma_twt^2}{D}[/tex]
Where
K=Constant
t=Thickness
D=Die opening
w=Width
[tex]\sigma_t[/tex]=Tensile strength
Here K= 0.33
Now by putting the all given values
[tex]F=K\dfrac{\sigma_twt^2}{D}[/tex]
[tex]F=0.33\times \dfrac{340\times 10^6\times 0.02\times (0.005)^2}{0.008}[/tex]
F=7012.5 N
F=7.012 KN
