Asked by Danny
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.
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.
Answers
Answered by
Elena
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
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
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