Question:-
Statement for Linked Answer Questions: 54 & 55

In a shear cutting operation, a sheet of 5mm thickness is cut along a length of 200mm.
The cutting blade is 400mm long and zero-shear (S=0) is provided on the edge. The
ultimate shear strength of the sheet is 100MPa and penetration to thickness ratio is 0.2. Neglect friction.



Q. Assuming force vs displacement curve to be rectangular, the work done (in J) is
Option (A)
100
Option (B)
200
Option(C)
250
Option(D)
300
Correct Option:
(B)
Question Solution:
Work Done = t x shear area x K.t

Where K= penetration =0.2; T= thickness= 5 mm

Shear area = (200+200)X5=2000 mm2

Therefore W = 100X2000X5X0.2 = 200000 N-mm = 200 J

Work Done with shear = FmaxX(Kt + I) = 200

Fmax = 200/(5x0.2+20) = 9.5 KN nearly 10KN
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