# Civil Engineering - Theory of Structures - Discussion

### Discussion :: Theory of Structures - Section 1 (Q.No.16)

16.

A load of 1960 N is raised at the end of a steel wire. The minimum diameter of the wire so that stress in the wire does not exceed 100 N/mm2 is :

 [A]. 4.0 mm [B]. 4.5 mm [C]. 5.0 mm [D]. 5.5 mm [E]. 6.0 mm

Answer: Option C

Explanation:

No answer description available for this question.

 Sanjeev said: (Sep 2, 2013) Stress = load/area.

 Jack said: (Oct 22, 2013) Given: Load = 1960N. Area of Wire = ? Allowable Stress = 100 N/mm2. So Stress = P/A. A= P/Stress. A= 1960N/100N/mm2. A= 19.6mm2. Dia = Pid2/4. 19.6mm2 = 3.14d2/4. (19.6*4/3.14)^.5 = d = 5.00mm.

 Arghya said: (Dec 22, 2014) The answer is 4.9 we have to take the approx value which is 5.00.

 Vikram Singh said: (Apr 27, 2016) Very good explanation @Jack.

 Krishna said: (Feb 13, 2017) Nice explanation. Thanks.

 Maitri said: (Jul 12, 2017) Thanks @Jack.

 Harsha said: (Jul 27, 2017) Stress = force/area. Area = force/stress, Area = 1960/100=19.6. πdd/4=19.6, dd/4=19.6/3.14, dd=6.24*4, d= √24.96, d=5mm.

 Tirusew S. said: (Oct 12, 2017) Thank you all for the given explanation.

 Seenu said: (Nov 25, 2017) 2*PI*R = 2*π*d/2*d/2.

 Monika said: (Oct 12, 2018) Thanks for the explanation.

 K.Vignesh. said: (Jan 20, 2019) Good explanation, Thanks @Jack.

 Rajiv said: (Sep 2, 2020) P= 1960 N. Stress= 100N/mm^2. Formula(Sress= P/A). 100 = 1960/A. A = 1960/100. Where A=π/4d^2. Then :- 22/7*4*d^2=196/10. d^2= 28/22*196/10. d= √25. d= 5 =>Answer.

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