Q#21
A steel wire of cross-sectional area 15 mm2 has an ultimate tensile stress of 4.5 × 108 N m-2.
(a) Calculate the maximum tension that can be applied to the wire. [2]
(b) The steel of the wire has density 7800 kg m-3. The wire is hung vertically.
Calculate the maximum length of the steel wire that could be hung vertically before the wire breaks under its own weight. [3]
Solution:
(a)
{Stress = Force / Area
Note that the tension is a force.}
stress σ = F / A
{Convert area into m2 by multiplying by 10-6.
Tension = Stress × Area}
max. tension = UTS × A = 4.5×108 × 15×10-6 = 6800 (6750) N
(b)
{We first need to relate the length with data available.}
Density ρ = m / V
weight = mg = ρVg = ρALg
{since Volume V = Area × Length = AL}
{When the wire break under its own weight, the weight should be equal to the maximum tension.
Weight = ρALg
Tension = ρALg}
6750 = 7.8×103 × 15×10-6 × L × 9.81
L = 5.9 (5.88) × 103 m
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