A steel wire of cross-sectional area 15 mm2 has an ultimate tensile stress of 4.5 × 108 N m-2.

Question 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]

 

 

 

Reference: Past Exam Paper – November 2015 Paper 23 Q7

 

 

 

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