If bar is subjected to an axial tensile load of 15 kN, find its total extension,

$$ \text { Assume: } E_{\text {steel}}=2 \times 10^{5} \mathrm{N} / \mathrm{mm}^{2} \quad \text { Find: } \delta_{L}=? $$ ![enter image description here][1] $$ \begin{aligned} \delta L &=\delta L_{1}+\delta L_{2}+\delta L_{3} \ \delta L &=\left(\frac{P L}{A E}\right)_{1}+\left(\frac{P L}{A E}\right)_{2}+\left(\frac{P L}{A E}\right)_{3} \ \delta L &=\frac{P}{E}\left(\frac{L_{1}}{A_{1}}+\frac{L_{2}}{A_{2}}+\frac{L_{3}}{A_{3}}\right) \end{aligned} $$ $$ \begin{aligned} \delta L=& \frac{P}{E\left(\frac{\pi}{4}\right)}\left(\frac{L_{1}}{d_{1}^{2}}+\frac{L_{2}}{d_{2}^{2}}+\frac{L_{3}}{d_{3}^{2}}\right) \ \delta L=& \frac{15 \times 10^{3}}{2 \times 10^{5}\left(\frac{\pi}{4}\right)}\left(\frac{200}{40^{2}}+\frac{500}{15^{2}}+\frac{200}{30^{2}}\right) \ \delta L=0.245 \mathrm{mm} \end{aligned} $$

$$ \text { Assume: } E_{\text {steel}}=2 \times 10^{5} \mathrm{N} / \mathrm{mm}^{2} \quad \text { Find: } \delta_{L}=? $$

$$ \text { Assume: } E_{\text {steel}}=2 \times 10^{5} \mathrm{N} / \mathrm{mm}^{2} \quad \text { Find: } \delta_{L}=? $$