Question Number 98575 by aurpeyz last updated on 14/Jun/20
$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{force}\:\mathrm{F}\:\mathrm{needed}\:\mathrm{to}\:\mathrm{punch}\:\mathrm{at}\:\mathrm{1}.\mathrm{46cm}\:\mathrm{diametre}\:\mathrm{hole}\:\mathrm{in}\:\mathrm{a}\:\mathrm{steel}\:\mathrm{plate}\:\mathrm{1}.\mathrm{27cm}\:\mathrm{tjick}.\:\mathrm{tbe}\:\mathrm{ultimate}\:\mathrm{shear}\:\mathrm{stress}\:\mathrm{of}\:\mathrm{the}\:\mathrm{steel}\:\mathrm{is}\:\mathrm{3}.\mathrm{45}×\mathrm{10}^{\mathrm{8N}/\mathrm{m}^{\mathrm{2}} } \\ $$
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
$${shear}\:{stress}\:\tau=\frac{{F}}{{pt}}\geqslant{shear}\:{strength} \\ $$$${F}={pt}\tau \\ $$$${p}={peremeter}\:{of}\:{hole} \\ $$$${t}={thickness}\:{of}\:{plate} \\ $$$${round}\:{hole}:\:{p}=\pi{d}\:{with}\:{d}={diameter} \\ $$$${square}\:{hole}:\:{p}=\mathrm{4}{d}\:{with}\:{d}={side}\:{length} \\ $$
Commented by mr W last updated on 14/Jun/20
$${F}=\pi×\mathrm{14}.\mathrm{6}×\mathrm{12}.\mathrm{7}×\mathrm{0}.\mathrm{345}=\mathrm{200}.\mathrm{97}\:{KN} \\ $$
Commented by mr W last updated on 14/Jun/20
Commented by aurpeyz last updated on 15/Jun/20
$$\mathrm{i}\:\mathrm{am}\:\mathrm{for}\:\mathrm{ever}\:\mathrm{grateful}.\:\mathrm{thanks}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{am}\:\mathrm{really}\:\mathrm{learning} \\ $$