The force requirement in blanking operation of a metal sheet is 10 kN. The thickness of sheet is T and the diameter of blanked part is D. For the same material and same conditions, if the diameter of blanked part is increased to 1.7 D and thickness of the sheet is reduced to 0.5 T, what will be the new blanking force required?

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BHEL Engineer Trainee Mechanical 24 Aug 2023 Official Paper

Answer 1

Concept:

Blanking force is given by:

\( F = \tau \times \text{Perimeter} \times \text{Thickness} \)

For a circular blank, perimeter is \( \pi D \) and thickness is \( T \).

Therefore, \( F = \tau \times \pi D \times T \)

Given:

Initial blanking force = 10 kN

Initial diameter = \( D \), Initial thickness = \( T \)

New diameter = \( 1.7D \), New thickness = \( 0.5T \)

Calculation:

New force,

\( F_{\text{new}} = \tau \times \pi \times (1.7D) \times (0.5T) = \tau \times \pi D T \times (1.7 \times 0.5) \)

\( F_{\text{new}} = F_{\text{old}} \times 0.85 = 10 \times 0.85 = 8.5~\text{kN} \)