How much force is required to punch a hole 1 cm in diameter in a steel sheet $5\;mm$ thick whose shearing strength is $2.76 \; * \; 10^8\;N/m^2$.

Given,
Diameter of the hole $(d) = 1\;cm = 0.01\;m$
Thickness of the sheet $(t) = 5\;mm = 5 * 10^{-3}\;m$
Shear of strength $(\eta) = 2.76 * 10^8 \;N/m^2$
Force required $(F) = ?$

We have,

Shear strength $= \frac{F}{A}$ .......... (i)

But $A = l * t = 2 . \pi . r * t$          [∵ Length of the circumference $l = 2 . \pi . r$]

From the equation $1^{st}$,

$F = Shear \;\; strength * A = 2.76 * 10^8 * 2 . \pi . r * t$

$\;\;\;\; = 2.76 * 10^8 * 2 . \pi . 0.01 * 5 * 10^{-3} = 43353\;N$

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