Discussion
A steel rod 1 m long and 20 mm × 20 mm in cross-section is subjected to a tensile force of 40 kN. What is the elongation of the rod, if the modulus of elasticity for the rod material is 200 GPa.
Answer: B
The elongation of an axially loaded bar is \(\delta = PL/(AE)\). The area is \(20 \times 20 = 400\) mm\(^2\), the load is 40,000 N, the length is 1000 mm and \(E = 200{,}000\) N/mm\(^2\):
\[ \begin{aligned} \delta \\ &= \frac{40000 \times 1000}{400 \times 200000} \\ &= 0.5\ \text{mm} \end{aligned} \]The options 0.25, 0.75 and 0.95 mm correspond to a wrong load, area or modulus (for example 0.25 mm would be found with an area of 800 mm\(^2\)).
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