written 5.2 years ago by | modified 2.6 years ago by |
$E_{c} = 11- GPa and Eal = 70 GPa$
written 5.2 years ago by | modified 2.6 years ago by |
$E_{c} = 11- GPa and Eal = 70 GPa$
written 5.2 years ago by |
$$ \begin{array}{l}{\text { Data: } \mathrm{Es}=210 \mathrm{kN} / \mathrm{mm}^{2}, \mathrm{Ec}=110 \mathrm{GPa} \text { and } \mathrm{E}_{\mathrm{Al}}=70 \mathrm{GPa}} \ {\text { Find: } \mathrm{P}, \delta \mathrm{L}} \ {\text { To find unknown force } \mathrm{P} \text { , }} \ {\sum \mathrm{Fx}=0} \ {-30+\mathrm{P}-5+10=0} \ {\mathrm{P}-25=0} \ {\mathrm{P}=25 \mathrm{kN}}\end{array} $$ To find forces acting on individual part of compound rod. ![enter image description here][1] $$ \begin{array}{l}{\delta \mathrm{L}=\delta \mathrm{L}_{1}+\delta \mathrm{L}_{2}+\delta \mathrm{L}_{3}} \ {\delta \mathrm{L}=\left(\frac{\mathrm{PL}}{\mathrm{AE}}\right)_{1}+\left(\frac{\mathrm{PL}}{\mathrm{AE}}\right)_{2}+\left(\frac{\mathrm{PL}}{\mathrm{AE}}\right)_{3}}\end{array} $$ $$ \begin{array}{l}{\delta \mathrm{L}=\left(\frac{30 \times 10^{3} \times 1200}{\frac{\pi}{4} \times 35^{2} \times 210 \times 10^{3}}\right)_{1}+\left(\frac{5 \times 10^{3} \times 1000}{\frac{\pi}{4} \times 20^{2} \times 110 \times 10^{3}}\right)_{2}+\left(\frac{10 \times 10^{3} \times 800}{10 \times 10 \times 70 \times 10^{3}}\right)_{3}} \ {\delta \mathrm{L}=0.1782+0.1447+1.1485} \ {\delta \mathrm{L}=1.466 \mathrm{mm}}\end{array} $$