The force required to punch a square hole of | vedclass

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Question

(A) The shearing force acts on the area of the boundary (perimeter) of the hole.The perimeter of the square hole is $P = 4 	imes 	ext{side} = 4 	im

Vedclass · Accepted Answer

$5.6 	imes 10^4\,N$

Vedclass · Answer

(A) The shearing force acts on the area of the boundary (perimeter) of the hole.The perimeter of the square hole is $P = 4 	imes 	ext{side} = 4 	imes 2\,cm = 8\,cm = 0.08\,m$.The thickness of the sheet is $t = 2\,mm = 0.002\,m = 2 	imes 10^{-3}\,m$.The area $A$ over which the shearing force acts is the product of the perimeter and the thickness:$A = P 	imes t = 0.08\,m 	imes 0.002\,m = 1.6 	imes 10^{-4}\,m^2$.Given the shearing stress $	au = 3.5 	imes 10^8\,N/m^2$, the force $F$ is calculated as:$F = 	au 	imes A$$F = (3.5 	imes 10^8\,N/m^2) 	imes (1.6 	imes 10^{-4}\,m^2)$$F = 5.6 	imes 10^4\,N$.

The force required to punch a square hole of side $2\,cm$ in a steel sheet of thickness $2\,mm$ is (given: shearing stress of steel sheet $= 3.5 \times 10^8\,N/m^2$)

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