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Question-159157




Question Number 159157 by physicstutes last updated on 13/Nov/21
Answered by mr W last updated on 13/Nov/21
Commented by mr W last updated on 13/Nov/21
let AA′=1  ⇒BB′=((50)/(450))×1=(1/9)  ⇒CC′=BB′=(1/9)  ⇒DD′=((75)/(300+75))×(1/9)=(1/(45))  P×1=9×(1/(45))  ⇒P=(1/5) KN  M_D =(1/5)×(450+50+300)  =160 KNmm=160 Nm
$${let}\:{AA}'=\mathrm{1} \\ $$$$\Rightarrow{BB}'=\frac{\mathrm{50}}{\mathrm{450}}×\mathrm{1}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\Rightarrow{CC}'={BB}'=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\Rightarrow{DD}'=\frac{\mathrm{75}}{\mathrm{300}+\mathrm{75}}×\frac{\mathrm{1}}{\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{45}} \\ $$$${P}×\mathrm{1}=\mathrm{9}×\frac{\mathrm{1}}{\mathrm{45}} \\ $$$$\Rightarrow{P}=\frac{\mathrm{1}}{\mathrm{5}}\:{KN} \\ $$$${M}_{{D}} =\frac{\mathrm{1}}{\mathrm{5}}×\left(\mathrm{450}+\mathrm{50}+\mathrm{300}\right) \\ $$$$=\mathrm{160}\:{KNmm}=\mathrm{160}\:{Nm} \\ $$
Commented by Tawa11 last updated on 13/Nov/21
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$

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