Question Number 31114 by rahul@ last updated on 02/Mar/18
Commented by rahul@ last updated on 02/Mar/18
$${plz}\:{solve}\:{it}\:..{its}\:{urgent} \\ $$
Commented by rahul@ last updated on 03/Mar/18
$${thank}\:{u}\:{sir} \\ $$
Answered by mrW2 last updated on 02/Mar/18
![force in alu and in steel=F final length=L cross section of alu rod=A_A cross section of steel rods=A_S =2A_A l_0 (1+α_A θ)−L=Fl_0 /A_A Y_A ...(i) L−l_0 (1+α_S θ)=Fl_0 /A_S Y_S ...(ii) (i)/(ii): ((l_0 (1+α_A θ)−L)/(L−l_0 (1+α_S θ)))=((A_S Y_S )/(A_A Y_A ))=((2Y_S )/Y_A ) 2LY_S −2l_0 (1+α_S θ)Y_S =l_0 (1+α_A θ)Y_A −LY_A L(2Y_S +Y_A )=[2(1+α_S θ)Y_S +(1+α_A θ)Y_A ]l_0 L=(([2(1+α_S θ)Y_S +(1+α_A θ)Y_A ]l_0 )/(2Y_S +Y_A )) L=(([2Y_S +2α_S θY_S +Y_A +α_A θY_A ]l_0 )/(2Y_S +Y_A )) ⇒L=[1+(((2α_S Y_S +α_A Y_A )θ)/(2Y_S +Y_A ))]l_0](https://www.tinkutara.com/question/Q31123.png)
$${force}\:{in}\:{alu}\:{and}\:{in}\:{steel}={F} \\ $$$${final}\:{length}={L} \\ $$$${cross}\:{section}\:{of}\:{alu}\:{rod}={A}_{{A}} \\ $$$${cross}\:{section}\:{of}\:{steel}\:{rods}={A}_{{S}} =\mathrm{2}{A}_{{A}} \\ $$$${l}_{\mathrm{0}} \left(\mathrm{1}+\alpha_{{A}} \theta\right)−{L}={Fl}_{\mathrm{0}} /{A}_{{A}} {Y}_{{A}} \:\:\:…\left({i}\right) \\ $$$${L}−{l}_{\mathrm{0}} \left(\mathrm{1}+\alpha_{{S}} \theta\right)={Fl}_{\mathrm{0}} /{A}_{{S}} {Y}_{{S}} \:\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\frac{{l}_{\mathrm{0}} \left(\mathrm{1}+\alpha_{{A}} \theta\right)−{L}}{{L}−{l}_{\mathrm{0}} \left(\mathrm{1}+\alpha_{{S}} \theta\right)}=\frac{{A}_{{S}} {Y}_{{S}} }{{A}_{{A}} {Y}_{{A}} }=\frac{\mathrm{2}{Y}_{{S}} }{{Y}_{{A}} } \\ $$$$\mathrm{2}{LY}_{{S}} −\mathrm{2}{l}_{\mathrm{0}} \left(\mathrm{1}+\alpha_{{S}} \theta\right){Y}_{{S}} ={l}_{\mathrm{0}} \left(\mathrm{1}+\alpha_{{A}} \theta\right){Y}_{{A}} −{LY}_{{A}} \\ $$$${L}\left(\mathrm{2}{Y}_{{S}} +{Y}_{{A}} \right)=\left[\mathrm{2}\left(\mathrm{1}+\alpha_{{S}} \theta\right){Y}_{{S}} +\left(\mathrm{1}+\alpha_{{A}} \theta\right){Y}_{{A}} \right]{l}_{\mathrm{0}} \\ $$$${L}=\frac{\left[\mathrm{2}\left(\mathrm{1}+\alpha_{{S}} \theta\right){Y}_{{S}} +\left(\mathrm{1}+\alpha_{{A}} \theta\right){Y}_{{A}} \right]{l}_{\mathrm{0}} }{\mathrm{2}{Y}_{{S}} +{Y}_{{A}} } \\ $$$${L}=\frac{\left[\mathrm{2}{Y}_{{S}} +\mathrm{2}\alpha_{{S}} \theta{Y}_{{S}} +{Y}_{{A}} +\alpha_{{A}} \theta{Y}_{{A}} \right]{l}_{\mathrm{0}} }{\mathrm{2}{Y}_{{S}} +{Y}_{{A}} } \\ $$$$\Rightarrow{L}=\left[\mathrm{1}+\frac{\left(\mathrm{2}\alpha_{{S}} {Y}_{{S}} +\alpha_{{A}} {Y}_{{A}} \right)\theta}{\mathrm{2}{Y}_{{S}} +{Y}_{{A}} }\right]{l}_{\mathrm{0}} \\ $$
Commented by Tinkutara last updated on 05/Mar/19
Sir,
why in (i) L is subtracted but in (ii) it is subtracted from L?