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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 = $1.05 \times 10^{5} N/mm^{2}$
written 6.2 years ago by gravatar for teamques10 teamques10 ★ 70k •   modified 6.2 years ago

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written 6.2 years ago by gravatar for teamques10 teamques10 ★ 70k

Given: For brass bar, $A = 1000 mm^{2} , E = 1.05 \times 10^{5} N/mm^{2} $

Solution:

enter image description here

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}} = …

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