Physics (Stress & Strain)

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A hole 3cm in diameter is to be punched out of a steel plate 8cm thick. The shear stress of the material is 670GPa. What load is required on the punch?
3cm = 0.03m
8cm = 0.08m
Lateral surface area
(As) = pi*D*h
= 3.1416 * .03m * .08m
= 3.1416 * .0024m
= .007539m
Force = Stress * Area
= 670GPa * .007539m
= 5.05113GN
= 5051.13MN
The load required is 5051.1MN.
Is this correct? Thank you.

  • Physics (Stress & Strain) -

    Shearing stress (tangential stress)
    τ=V/A,
    where V ia resultant shearing force which passes through the area A being sheared.
    V= τA=τ•πdh
    Shear force is equal to the punching force P
    P=τπdh=
    =670•10⁹•3.14•0.03•0.08 =
    =5.05•10⁹Pa =5.05GPa

  • Physics (Stress & Strain) -

    That is correct.

    perimeter of circle x thickness
    (pi x d) x 0.08m
    pi x 0.03 m x 0.08 m = 0.0075396 m2

    stress = load / area
    load = stress x area = 670 GPa x 0.0075396 m2
    = 670 GN/m2 x 0.0075396 m2
    = 5.051532 GN or 5051.532 MN

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