QuestionA steel cable of length 20 m and cross-sectional area 2.0 × 10-4 m2 supports a load of 3.0 × 104 N. For this steel the Young modulus is 2.0 × 1011 Pa and the yield stress is 2.5 × 108 Pa.(a) Calculate the stress in the cable. (1 mark)
(b) Explain whether the cable will return to its original length when the load is removed. (2 marks)
(c) Calculate the extension of the cable. (3 marks)
(d) Calculate the maximum load the cable can support before it yields. (2 marks)Worked answer(a) Stress:σ = F ÷ A = 3.0 × 104 ÷ 2.0 × 10-4 = 1.5 × 108 Pa (b) 1.5 × 108 Pa is well below the yield stress, so the steel is still deforming elastically and the cable will return to its original length.(c) Rearrange E = σ ÷ ε to find the strain, then the extension:ε = σ ÷ E = 1.5 × 108 ÷ 2.0 × 1011 = 7.5 × 10-4
Δx = ε × x = 7.5 × 10-4 × 20 = 0.015 m = 15 mm (d) Maximum load:F = σ × A = 2.5 × 108 × 2.0 × 10-4 = 5.0 × 104 N Examiner tip: in (b) you must compare the stress with the yield stress and use the comparison to justify your answer; just writing "yes" scores nothing. In (c) remember that strain has no unit, so the extension comes out in the same unit as the original length.
(b) Explain whether the cable will return to its original length when the load is removed. (2 marks)
(c) Calculate the extension of the cable. (3 marks)
(d) Calculate the maximum load the cable can support before it yields. (2 marks)Worked answer(a) Stress: