written 5.2 years ago by
teamques10
★ 68k
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Given: For brass bar, $A = 1000 mm^{2} , E = 1.05 \times 10^{5} N/mm^{2} $
Solution:
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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