Question

A hole of 50 mm diameter is to be punched on a 2 mm thick sheet. The shear stress for the material can be taken as 3.1 $kN/m{m}^2$. Assuming gradual variation of load with punch travel and penetration to thickness ratio = 0.5, the energy required to punch the hole is

• A. 973.89 J
• B. 1216.21 J
• C. 1947.78 J
• D. 2609.76 J

Answer 1

The correct option is A 973.89 J

$\text{Punching force}\left(F\right)=\tau \left(\pi D\right)\times t$

$=3.1\times 1000\times \pi \times 50\times 2$

$=973.894kN$

$\therefore$ Energy required

$=\frac{1}{2}\times F\times t=973.89J$