Question: A brass bar having cross sectional area of $1000^{2}$ is subjected to axial forces s shown in Fig.no 3, Find the net deformation in the bar . Take E = $1.05 \times 10^{5} N/mm^{2}$
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modified 8 days ago  • written 8 days ago by gravatar for bharathchippa49 bharathchippa490
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Given: For brass bar, $A = 1000 mm^{2} , E = 1.05 \times 10^{5} N/mm^{2} $

Solution:

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1) $\sigma_{LAB} = \big(\frac{PL}{AE}\big)_{AB} = \frac{50 \times 10^{3} \times 800}{1000 \times 1.05 \times 10^{5}} = + 0.381 mm$

2) $\sigma_{LBC} = \big(\frac{PL}{AE}\big)_{BC} = \frac{30 \times 10^{3} \times 1000}{1000 \times 1.05 \times 10^{5}} = + 0.286 mm$

3) $\sigma_{LCD} = \big(\frac{PL}{AE}\big)_{CD} = \frac{10 \times 10^{3} \times 1200}{1000 \times 1.05 \times 10^{5}} = + 0.114 mm$

Net deformation = $\sigma_{LAB} + \sigma_{LBC} + \sigma_{LCD}$

= 0.381 - 0.286 + 0.114

$\sigma L = - 0.019 mm $ (negative sign indicate decrease in length)

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written 8 days ago by gravatar for bharathchippa49 bharathchippa490
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