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A steel bar is 900mm long; its two ends are 40mm and 30mm in diameter and the length of each part of rod is 200mm. The middle portion of the bar is 15 mm in diameter and 500 mm long.

written 5.5 years ago by teamques10 ★ 69k

modified 2.9 years ago by RakeshBhuse • 3.2k

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

strength of materials

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written 5.5 years ago by teamques10 ★ 69k

•   modified 5.5 years ago

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

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written 5.5 years ago by teamques10 ★ 69k

•   modified 5.5 years ago

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

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written 5.5 years ago by teamques10 ★ 69k

•   modified 5.5 years ago

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

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