Answer to Question #153224 in Mechanical Engineering for Amrish

Question

Determine the smallest size of hole that can be punched in 12 mm thick MS 

plate having ultimate shear stress of 390 N/mm2 , if permissible crushing 

stress for the punch material is 2.4 kN/mm2.

Answer 1

Ultimate shear stress=$\tau_u=$ 390 N/mm², thickness= 12mm,

Permissible crushing stress=2.4 kN/mm²

Axial shear force= $(\frac{\pi d^2}{4})\tau= F$ ,In case of crushing

$F=\pi dt \times \sigma_c$ as we know that in case of small size both force will be same

$\pi dt \sigma_c=\frac{\pi d^2}{4}\times \tau_u$

12 $\times 2400= \frac{d}{4}\times 390$

d=295.38 mm