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Question Number 98575 by aurpeyz last updated on 14/Jun/20

$$\mathrm{Calculate}\mathrm{the}\mathrm{force}\mathrm{F}\mathrm{needed}\mathrm{to}\mathrm{punch}\mathrm{at}1.46\mathrm{cm}\mathrm{diametre}\mathrm{hole}\mathrm{in}\mathrm{a}\mathrm{steel}\mathrm{plate}1.27\mathrm{cm}\mathrm{tjick}.\mathrm{tbe}\mathrm{ultimate}\mathrm{shear}\mathrm{stress}\mathrm{of}\mathrm{the}\mathrm{steel}\mathrm{is}3.45\times {10}^{8\mathrm{N}/{\mathrm{m}}^2}$$

Commented by aurpeyz last updated on 14/Jun/20

$$\mathrm{pls}\mathrm{he}\mathrm{me}\mathrm{with}\mathrm{it}?$$

Commented by mr W last updated on 14/Jun/20

Commented by mr W last updated on 14/Jun/20

$$shearstress\tau =\frac{F}{pt}⩾shearstrength$$$$F=pt\tau$$$$p=peremeterofhole$$$$t=thicknessofplate$$$$roundhole:p=\pi dwithd=diameter$$$$squarehole:p=4dwithd=sidelength$$

Commented by mr W last updated on 14/Jun/20

$$F=\pi \times 14.6\times 12.7\times 0.345=200.97KN$$

Commented by mr W last updated on 14/Jun/20

Commented by aurpeyz last updated on 15/Jun/20

$$\mathrm{i}\mathrm{am}\mathrm{for}\mathrm{ever}\mathrm{grateful}.\mathrm{thanks}\mathrm{sir}.\mathrm{i}\mathrm{am}\mathrm{really}\mathrm{learning}$$

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