Question #153224

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.


1
Expert's answer
2021-01-06T06:04:12-0500

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


Permissible crushing stress=2.4 kN/mm2


Axial shear force= (πd24)τ=F(\frac{\pi d^2}{4})\tau= F ,In case of crushing


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


πdtσc=πd24×τu\pi dt \sigma_c=\frac{\pi d^2}{4}\times \tau_u


12 ×2400=d4×390\times 2400= \frac{d}{4}\times 390


d=295.38 mm


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS