Civil Engineering - Theory of Structures - Discussion

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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