Question Number 98568 by aurpeyz last updated on 14/Jun/20
Commented by aurpeyz last updated on 14/Jun/20
$$\mathrm{pls}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{question}\:\mathrm{13}.\mathrm{9} \\ $$
Answered by mr W last updated on 14/Jun/20
$${thickness}\:{t}=\mathrm{2}.\mathrm{5}{mm} \\ $$$${diameter}\:{of}\:{hole}\:{d}=\mathrm{6}{mm} \\ $$$${shear}\:{stress}\:=\frac{{F}}{\pi{dt}}=\mathrm{3}.\mathrm{0}×\mathrm{10}^{\mathrm{8}} \:{Pa}=\mathrm{0}.\mathrm{3}\:\frac{{KN}}{{mm}^{\mathrm{2}} } \\ $$$$\Rightarrow{F}=\mathrm{0}.\mathrm{3}×\pi×\mathrm{6}×\mathrm{2}.\mathrm{5}=\mathrm{14}.\mathrm{14}\:{KN} \\ $$$$ \\ $$$${if}\:{the}\:{hole}\:{is}\:{a}\:{square},\:{then} \\ $$$$\Rightarrow{F}=\mathrm{0}.\mathrm{3}×\mathrm{4}×\mathrm{6}×\mathrm{2}.\mathrm{5}=\mathrm{18}\:{KN} \\ $$
Commented by aurpeyz last updated on 15/Jun/20
$$\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}.\:\mathrm{thanks}\:\mathrm{sir} \\ $$