Topic: Strength of Materials
A 20 kN weight is suspended by two wires as shown in the figure. The length of each wire is 2 meter. The steel wire (E value 200GPa) has an cross-sectional area of $60\times10^{-6}m^{2}$ and the aluminium wire (E value 70GPa) has an cross-sectional area of $120\times10^{-6}m^{2}$. The stress in aluminium wire is
- A. $\frac{17}{102}GPa$
- B. $\frac{7}{120}GPa$
- C. $\frac{17}{120}GPa$
- D. $\frac{7}{102}GPa$
Answer: D. $\frac{7}{102}GPa$
Explanation: Since the wires are in parallel, their deflections are equal: $\delta_{Al} = \delta_{st}$. Therefore, $\frac{P_{Al}L_{Al}}{A_{Al}E_{Al}} = \frac{P_{st}L_{st}}{A_{st}E_{st}}$. Substituting the given values: $\frac{P_{Al}}{P_{st}} = \frac{120\times10^{-6} \times 70}{60\times10^{-6} \times 200} = \frac{7}{10}$. So, $P_{st} = \frac{10}{7}P_{Al}$. From equilibrium, $P_{Al} + P_{st} = 20$ kN. Thus, $P_{Al} + \frac{10}{7}P_{Al} = 20$, which gives $P_{Al} = \frac{140}{17}$ kN. The stress in the aluminium wire is $\sigma_{Al} = \frac{P_{Al}}{A_{Al}} = \frac{140\times10^{3}}{17 \times 120\times10^{-6}} = \frac{7}{102} \times 10^{9}$ N/m² = $\frac{7}{102}$ GPa.