Question

The punching force required in a blanking operation of mild steel sheet is 500 kN. The diameter of blank is increased by 20% and thickness is reduced by 4%, then punching force will be:

• A. 434 kN
• B. 576 kN
• C. 634 kN
• D. 676 kN

Answer 1

The correct option is B 576 kN

$F=\pi dt\tau$

$500=\pi dt\left(\tau \right)$ ...(i)

$F=\pi \times \left(1.2d\right)\times \left(0.96t\right)\tau$ ...(ii)

Dividing (i) and (ii)

$\frac{500}{F}=\frac{\pi dt\tau }{\pi \times \left(1.2d\right)\times \left(0.96t\right)\tau }$

$F=576kN$