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The force requirement in a blanking operation of low carbon steel sheet is 5.0 kN. The thickness of the sheet is 't' and diameter of the blanked part is 'd'. For the same work material, if the diameter of the blanked part is increased to 1.5d and thickness is reduced to 0.4t, the new blanking force in kN is
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- 3.0
- 4.5
- 6.7
- 0.8
Correct Option: A
Let τ be the shear stress
F = t × d × t = 5.0 kN
F1 = r × 1.5 d × 4 × t = .6 × 5 = 3 kN