Question

A hole 75 mm is to be punched in a steel plate 5.6 mm thick. The ultimate shear stress is $500N/m{m}^2$. It cutting is complete at 40 percent penetration of the punch and shear of 2 mm is provided on the punch , The minimum force required in the punch is ____kN

• A. 348.538

Answer 1

The correct option is A 348.538
$F_{\text{max}}=Lt\tau$
$=(\pi \times D)\times t\times \tau$
$=(\pi \times 75)\times 5.6\times 500=659.734\text{kN}$
$P_t=0.40\times 5.6mm=2.24mm$

$F=\frac{F_{\text{max}}\times Pt}{s+Pt}$

Shear on punch and it is comparable,
$F_{\text{min}}=\frac{659.734\times 0.40\times 5.6}{2+2.24}=348.538\text{kN}$