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A compound bar having steel rod of diameter 35mm and solid copper of diameter 20mm and aluminiu square rod is as shown in fig. Find change in length of bar. Take Modullus of elasticity $E_{s}=210$

$E_{c} = 11- GPa and Eal = 70 GPa$

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$$ \begin{array}{l}{\text { Data: } \mathrm{Es}=210 \mathrm{kN} / \mathrm{mm}^{2}, \mathrm{Ec}=110 \mathrm{GPa} \text { and } \mathrm{E}_{\mathrm{Al}}=70 \mathrm{GPa}} \ {\text { Find: } \mathrm{P}, \delta \mathrm{L}} \ {\text { To find unknown force } \mathrm{P} \text { , }} \ {\sum \mathrm{Fx}=0} \ {-30+\mathrm{P}-5+10=0} \ {\mathrm{P}-25=0} \ {\mathrm{P}=25 \mathrm{kN}}\end{array} $$ To find forces acting on individual part of compound rod. ![enter image description here][1] $$ \begin{array}{l}{\delta \mathrm{L}=\delta \mathrm{L}_{1}+\delta \mathrm{L}_{2}+\delta \mathrm{L}_{3}} \ {\delta \mathrm{L}=\left(\frac{\mathrm{PL}}{\mathrm{AE}}\right)_{1}+\left(\frac{\mathrm{PL}}{\mathrm{AE}}\right)_{2}+\left(\frac{\mathrm{PL}}{\mathrm{AE}}\right)_{3}}\end{array} $$ $$ \begin{array}{l}{\delta \mathrm{L}=\left(\frac{30 \times 10^{3} \times 1200}{\frac{\pi}{4} \times 35^{2} \times 210 \times 10^{3}}\right)_{1}+\left(\frac{5 \times 10^{3} \times 1000}{\frac{\pi}{4} \times 20^{2} \times 110 \times 10^{3}}\right)_{2}+\left(\frac{10 \times 10^{3} \times 800}{10 \times 10 \times 70 \times 10^{3}}\right)_{3}} \ {\delta \mathrm{L}=0.1782+0.1447+1.1485} \ {\delta \mathrm{L}=1.466 \mathrm{mm}}\end{array} $$

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