Self-Weight Elongation and Maximum Hanging Length
For a uniform vertical member of density rho and area A, local stress at distance x above the lower end is rho g x, giving total self-extension rho g L squared divided by 2Y.
Why this shows up in the exam
Long cables and mine ropes · Hanging rods · Maximum unsupported wire length
Learn the idea
A hanging member's tension varies with position because each section supports the material below it. The top of a hanging wire carries all the wire below while the bottom carries almost none, so strain is not uniform and must be accumulated along the length.
🧠 Memory hook: The top carries the whole hanging length; average tension is half the top value.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- Delta L_self = rho g L²/(2Y) — extension of a uniform hanging member under its own weight
- sigma_top = rho g L — maximum self-weight stress at the support
- L_max = sigma_break/(rho g) — maximum ideal hanging length before top-section failure
How to approach it
- 1Write the load below a general section
- 2Convert local stress to local strain
- 3Integrate for extension or use top stress for failure
Common slip-ups that cost marks
- •Using Mg uniformly along the wire
- •Including area in L_max when it cancels
- •Missing the factor one-half in self-extension
🌟 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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