Examples:$-$ Compound Bar's:$-$

$R_0$

$($a$)$ obbain ${}_A$ and ${}_s$ in stael $ Aluminiom.

$($b$)$ Calculate stresses for $12kN$ force and Cross section of $10m{m}^2$ in Aluninic and $2\mathrm{.}5m{m}^2$ in steel Modulus of Elasticity is to GPA for AL and $210$ Gipa for steel.

Sof:$-$: $,{\displaystyle \underset{?}{\overset{?}{}}}{F}_y=0,R_A+R_C=r$

Compatilititip Equations:$-$

${}_{AC}={}_{BC}or{}_{AC}-{}_{BC}=0$

${}_{AC}=\frac{R_A*L}{A_S{E}_S}$ and ${}_{BC}=\frac{R_C*2L}{A_A{E}_A}$

Substituting in $(2)$

$\frac{R_A*L}{A_S{E}_S}-\frac{R_C*2L}{A_A*E_A}=0$

From $(1),$ we get. $R_A=2P-R_C,$ substituting in $(3),$

$\frac{(2P-R_C)L}{A_S{E}_S}-\frac{R_C*2L}{A_A{E}_A}=0$

$o{r}_L(\frac{(2P-R_C)}{A_S{E}_S})=\frac{R_c*2L}{A_A{E}_A}$

$\frac{L(2P-R_C)}{2L*R_C}=\frac{A_{S}ES}{A_A{E}_A}=>\frac{4(ASES)}{A_A{E}_A}=\frac{2P}{R_C}-\frac{1}{8}=>\frac{4(A_{S}E3)+1}{A_A{E}_A}=\frac{P}{R_C}$

Rearranging, we get,

$\{\begin{array}{c}\text{:}{R}_C=\frac{P(A_A{E}_A)}{4(A_S{E}_S)+A_A{E}_N}\end{array}$