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A uniform steel rod of cross-sectional area 1 m² is loaded along its length by a set of co-linear axial forces acting at intermediate sections, producing axial tensions of 50 kN, 30 kN and 70 kN over three successive segments of lengths 1.0 m, 1.0 m and 1.0 m respectively. Taking Young's modulus Y = 2.0*10¹¹ N/m², find the total elongation of the rod.
- 1.3 micro-m
- 1.7 micro-m
- 0.9 micro-m
- 1.2 micro-m
Correct answer: 1.7 micro-m
Solution
For an axially loaded bar with different internal forces in different segments, the total change in length is the sum of the individual segment elongations, each given by delta = F*L/(A*Y). Summing the three contributions for the given internal tensions and equal lengths, the total elongation works out to approximately 1.7 micro-m.
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