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A 0.5 " ×0.5 " square is to be blanked from a 0.050 " thick piece of sheet metal. For the sheet material, the yield strength is 40ksi, tensile strength is 65ksi and fracture strength is 60ksi. The value of Ac​=0.05.

a) What force in tons will be required to shear all 3 parts simultaneously assuming a flat-bottomed punch?

1 Answer

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Final answer:

The force required to shear a 0.5" × 0.5" square from a 0.050" thick piece of sheet metal with a yield strength of 40ksi is approximately 1.2 tons.

Step-by-step explanation:

To calculate the force required to shear a 0.5" × 0.5" square from a sheet of metal, with the relevant mechanical properties given, we need to consider the shear strength of the material.

Typically, for ductile materials, the shear strength can be approximated as 0.6 times the yield strength, unless otherwise provided.

The shear force required (F) can be calculated by multiplying the shear strength (τ) by the area (A) being sheared:

F = τ × A

In our case, A is the perimeter of the square times the thickness of the sheet.

The calculation would be as follows:

  • A = 2(l + w) × thickness
  • A = 2(0.5" + 0.5") × 0.050"
  • A = 2 × 1" × 0.050"
  • A = 0.1 in²
  • τ = 0.6 × 40 ksi = 24 ksi
  • F = 24 ksi × 0.1 in²
  • F = 2.4 kips

To convert kips to tons, note that 1 kip = 0.5 tons, hence:

Force in tons = F / 2 = 2.4 kips / 2

= 1.2 tons

This is the theoretical force required, assuming perfect conditions and no friction.

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