and 4 others joined a min ago.
0
31kviews
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 = $1.05 \times 10^{5} N/mm^{2}$
1 Answer
5
8.1kviews
written 4.7 years ago by
teamques10
★ 65k
|
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}} = …
and 2 others joined a min ago.
ADD COMMENT
EDIT
Please log in to add an answer.