Civil Engineering - Theory of Structures - Discussion

25. 

The stiffness of the close coil helical spring is

[A].
[B].
[C].
[D].

Answer: Option A

Explanation:

No answer description available for this question.

P.Mohan said: (Aug 9, 2015)  
Explain how to derive it?

Mohan Kumar said: (Jan 27, 2017)  
We know deflection of spring= 64 wR^3N/cd^4.
Stiffness = load/ deflection.
So, subtitute R=D/2 & calculate stiffness, we get cd^4/8D^3n.

Murali said: (Aug 25, 2017)  
Explain the terms.

Bairu Ganesh said: (Oct 14, 2021)  
Deflection= 64w (r^3) n/Nd^4 where stiffness= load/deflection load p = w.

We know that radius r = D/2 and substitute i.e. deflection = 64w (D/2) ^3n/Nd^4=64wD^3n/8Nd^4.

= 8wD^3n/Nd^4.

Now stiffness = load/deflection = w/ (8wD^3n/Nd^4) = nd^4N/8D^3n.

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