Question

Circular blanks of $10$ mm diameter are punched from an aluminium sheet of $2$ mm thickness. The shear strength of aluminum is $80$ MPa. The minimum punching force required in kN is

• A. $2.57$
  
• B. $3.29$
• C. $5.03$
• D. $6.33$

Answer 1

The correct option is C $5.03$

Diameter of blank, $d=10$ mm
thickness of sheet, $t=2$ mm
Shear strength of aluminium sheet $\tau =80$ MPa
The punching force, $F=\tau \times \left(\pi dt\right)$
$=80\times \pi \times 10\times 2=5.023kN$