Question Number 210180 by BHOOPENDRA last updated on 02/Aug/24
Commented by mr W last updated on 02/Aug/24
$${is}\:{the}\:{load}\:{acting}\:{in}\:{the}\:{shear}\:{center} \\ $$$${in}\:{both}\:{cases}? \\ $$
Commented by BHOOPENDRA last updated on 02/Aug/24
$${Yes}\:{we}\:{can}\:{assume}\:{that} \\ $$
Answered by mr W last updated on 02/Aug/24
$${Querschnitt}\:\mathrm{1}: \\ $$$${I}=\frac{{t}\left(\mathrm{4}{a}\right)^{\mathrm{3}} }{\mathrm{12}}+\mathrm{2}\left(\mathrm{2}{a}\right){t}\left(\mathrm{2}{a}\right)^{\mathrm{2}} =\frac{\mathrm{40}{a}^{\mathrm{3}} {t}}{\mathrm{3}} \\ $$$$\tau_{{A}_{\mathrm{1}} } =\frac{{V}\left(\mathrm{4}{a}\right)\left({a}/\mathrm{2}\right)}{\frac{\mathrm{40}{a}^{\mathrm{3}} {t}}{\mathrm{3}}}=\frac{\mathrm{3}{V}}{\mathrm{20}{at}} \\ $$$$\tau_{{B}_{\mathrm{1}} } =\frac{{V}\left(\mathrm{2}{a}\right)\left(\mathrm{2}{a}\right)}{\frac{\mathrm{40}{a}^{\mathrm{3}} {t}}{\mathrm{3}}}\left(\mathrm{1}+\frac{\mathrm{2}{a}}{\mathrm{2}{a}}\right)=\frac{\mathrm{3}{V}}{\mathrm{5}{at}} \\ $$$${Querschnitt}\:\mathrm{2}: \\ $$$${I}=\frac{{t}\left(\mathrm{4}{a}\right)^{\mathrm{3}} }{\mathrm{12}}+\mathrm{2}\left({a}\right){t}\left(\mathrm{2}{a}\right)^{\mathrm{2}} =\frac{\mathrm{28}{a}^{\mathrm{3}} {t}}{\mathrm{3}} \\ $$$$\tau_{{A}_{\mathrm{2}} } =\frac{{V}\left(\mathrm{4}{a}\right)\left({a}/\mathrm{2}\right)}{\frac{\mathrm{28}{a}^{\mathrm{3}} {t}}{\mathrm{3}}}=\frac{\mathrm{3}{V}}{\mathrm{14}{at}}\:>\:\tau_{{A}_{\mathrm{1}} } \\ $$$$\tau_{{B}_{\mathrm{2}} } =\frac{{V}\left(\mathrm{2}{a}\right)\left({a}\right)}{\frac{\mathrm{28}{a}^{\mathrm{3}} {t}}{\mathrm{3}}}\left(\mathrm{1}+\frac{\mathrm{2}{a}}{{a}}\right)=\frac{\mathrm{9}{V}}{\mathrm{14}{at}}\:>\:\tau_{{B}_{\mathrm{1}} } \\ $$$$\Rightarrow{only}\:{b}\:{is}\:{true}. \\ $$
Commented by BHOOPENDRA last updated on 02/Aug/24
$${Thanks}\:{sir}\: \\ $$
Commented by mr W last updated on 02/Aug/24
Commented by BHOOPENDRA last updated on 02/Aug/24
$${Can}\:{you}\:{please}\:{tell}\:{Mr}.{W}\:{which}\:{formula}\: \\ $$$${you}\:{used}\:{to}\:{calculating}\:{moment}\:{of}\: \\ $$$${interia} \\ $$
Commented by mr W last updated on 02/Aug/24
Commented by mr W last updated on 02/Aug/24
$${for}\:{thin}−{walled}\:{cross}\:{section}: \\ $$$${t}\ll{b},\:{h} \\ $$$${part}\:\mathrm{1}:\:{I}=\frac{{th}^{\mathrm{3}} }{\mathrm{12}} \\ $$$${part}\:\mathrm{2}\:\&\:\mathrm{3}:\:{I}=\mathrm{2}×{bt}×\left(\frac{{h}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Sigma:\:{I}=\frac{{h}^{\mathrm{3}} {t}}{\mathrm{12}}+\frac{{bth}^{\mathrm{2}} }{\mathrm{2}} \\ $$
Commented by BHOOPENDRA last updated on 02/Aug/24
$${Thankyou}\:{sir}\: \\ $$$${means}\:{I}_{{total}} =\Sigma\left(\bar {{I}}_{{i}} +{A}_{{i}} {d}_{{i}} ^{\mathrm{2}} \right) \\ $$$${Where}\:{I}_{{total}\:} {moment}\:{of}\:{inertia}\:{of}\:{composite} \\ $$$${section}\: \\ $$$${and}\:{I}_{{i}\:} ^{ } {is}\:{the}\:{moment}\:{of}\:{inertia}\:{of} \\ $$$${individual}\:{segment}\:{about}\:{its}\:{own} \\ $$$${centroidal}\:{axis} \\ $$$${A}_{{i}} \:{is}\:{are}\:{of}\:{the}\:{individul}\:{segment} \\ $$$${and}\:{d}_{{i}} {the}\:{verticle}\:{distance}\:{from} \\ $$$${centroid}\:{of}\:{the}\:{segment}\:{to}\:{the}\: \\ $$$${neutral}\:{axis}\left(\boldsymbol{{NA}}\right) \\ $$$${Right}\:{Mr}.{W}\:{sir}? \\ $$
Commented by mr W last updated on 02/Aug/24
$${yes}.\:{for}\:{part}\:\mathrm{2}\:\&\:\mathrm{3}:\:{I}_{{i},\mathrm{0}} =\frac{{bt}^{\mathrm{3}} }{\mathrm{12}}\approx\mathrm{0}\:{due}\:{to} \\ $$$${very}\:{small}\:{thickness}\:{t}. \\ $$
Commented by BHOOPENDRA last updated on 02/Aug/24
$${Thankyou}\: \\ $$