Discussion :: Theory of Structures - Section 1 (Q.No.25)
The stiffness of the close coil helical spring is
Answer: Option A
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.
Now stiffness = load/deflection = w/ (8wD^3n/Nd^4) = nd^4N/8D^3n.
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