The figure below shows a steel rod of 50 mm^{2} cross sectional area. It is loaded at four points. K, L, M and N. Assume E_{steel} = 200 GPa. The total change in length of the rod due to loading is

- 1 μm
- -10 μm
- 16 μm
- -5 μm

Option 4 : -5 μm

Free

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

Free body diagram

Total change in length

\(\begin{array}{l} = \frac{{{P_1}{L_1}}}{{AE}} + \frac{{{P_2}{L_2}}}{{AE}} + \frac{{{P_3}{L_3}}}{{AE}}\\ = \frac{1}{{50 \times 200 \times {{10}^3}}}\left[ {100 \times 500 - 150 \times 800 + 50 \times 400} \right]\\ = - \frac{1}{{{{10}^7}}} \times 5 \times {10^4} = - 5\times {10^{-3}}mm = -5\mu m \end{array}\)

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