Evaluate Elongation of wire:

An aerial copper wire (E =1 x 105 N/mm²) 40 m long is having cross sectional area of 80 mm² and weighs 0.6 N per meter run. If wire is suspended vertically, calculate

(a) Elongation of wire because of self weight,

(b) Total elongation when a weight of 200 N is attached to the lower end of it, and

(c) Maximum weight which this wire can support at the lower end if limiting value of stress is 65 N/mm².

Sol.:  (a) Weight of wire W = 0.6 × 40 = 24 N The elongation because of self weight is,

δL =   L2/2 E; where ω  is the specific weight (weight per unit volume)

In the terms of total weight W =       ωAL

δL = WL²/E = 24 × (40 × 103)/2 × 80 × (1 × 105)

= 0.06 mm                                                                               .......ANS

(b) Extension because of weight P attached at lower end,

δL = PL/2E

= 200 × (40 × 10³)/80 × (1 × 10⁵)

= 1.0 mm

Total elongation of wire = 0.06 + 1.0 = 1.06 mm

(c) Maximum limiting stress = 65 N/ mm²

Stress because of self weight equals that produced by a load of half its weight applied at the end. That is

Stress because of self weight = (W/2)/A = (24/2)/80 = 0.15 N/ mm²

Remaining stress = 65 - 0.15 = 64.85 N/mm²

Maximum weight that the wire can support = 64.85 × 80 = 5188 N