Composite and Segmented Wires
For axially connected linear-elastic segments, compatibility gives total extension as the algebraic sum of segment extensions, Delta L_total = sum(F_i L_i/(A_i Y_i)).
Why this shows up in the exam
End-to-end wires · Layered hanging loads · Composite support members
Learn the idea
Segment extensions add, while each segment must use its own tension, length, area, and modulus. A joined wire is several elastic springs: series pieces carry the relevant transmitted force and their length changes add, but hanging loads can make the transmitted force differ by segment.
🧠 Memory hook: Find each section force, stretch each section, then add.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- Delta L_total = sum_i F_i L_i/(A_i Y_i) — extension of piecewise-uniform axial members
- 1/k_series = sum_i 1/k_i — equivalent compliance when the same force crosses every segment
- F_i = load supported below section i — section-force rule for vertical segmented wires
How to approach it
- 1Cut each segment and determine its internal force
- 2Compute every segment's FL over AY
- 3Add compatible extensions or compare the requested segments
Common slip-ups that cost marks
- •Assigning the same tension when intermediate loads exist
- •Averaging moduli without compliance weighting
- •Adding strains instead of actual extensions
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A wire 1 m long and cross-sectional area 2 mm^2 extends by 1 mm under a 200 N load. Find Young modulus.
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