Direction: In shear cutting operation, a sheet of 5 mm thickness is cut along a length of 200 mm. The cutting blade is 400 mm long (see figure) and zero-shear (S = 0) is provided on the edge. The ultimate shear strength of the sheet is 100 MPa and penetration to thickness ratio is 0.2. Neglect friction. 
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A shear of 20 mm {S = 20 mm) is now provided on the blade. Assuming force vs displacement curve to be trapezoidal, the maximum force (in kN) exerted is
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- 5
- 10
- 20
- 40
- 5
Correct Option: D
Length = 200 mm , Blade length = 400 mm , Zero shear (s = 0)
Ultimate shear strength of sheet = 100 MPa
| = 0.2 = k | ||
| Thickness |
∴ Shear area = (200 + 200) 5 = 2000 mm2
Now Work done = τ × shear area × k.t.
Work done = 100 × 2000 × 5 × 0.2 = 200 Joules
| ∴ Maximum force = | ||
| (kt + shear) |
| Maximum force = | = 9.5 ≈ 10 kN | |
| (5 × 0.2 + 20) |
