Topic: Strength of Materials
The figure shows an Aluminium rod of $25~mm^{2}$ cross sectional area. It is loaded at four points, K, L, M and N. Assume $E=67$ GPa for Aluminium. The total change in length of the rod due to loading as shown in close to
- A. $30~\mu m$
- B. $-10~\mu m$
- C. $-30~\mu m$
- D. $10~\mu m$
Answer: C. $-30~\mu m$
Explanation: Forces in sections: $P_{KL}=100~N$ (Tension), $P_{LM} = 100-250 = -150~N$ (Compression), $P_{MN} = 50~N$ (Tension). Total change in length $\Delta L = \frac{1}{AE}(P_{KL}L_{KL} + P_{LM}L_{LM} + P_{MN}L_{MN}) = \frac{1}{25\times10^{-6}\times67\times10^{9}}(100\times0.5 - 150\times0.8 + 50\times0.4) = \frac{1}{1675000}(50 - 120 + 20) = \frac{-50}{1675000} = -29.85\times10^{-6}~m \approx -30~\mu m$.