Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 1 (Q.No. 25)
25.
The stiffness of the close coil helical spring is
Discussion:
4 comments Page 1 of 1.
Bairu ganesh said:
3 years ago
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.
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.
(2)
Murali said:
7 years ago
Explain the terms.
Mohan kumar said:
8 years ago
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.
Stiffness = load/ deflection.
So, subtitute R=D/2 & calculate stiffness, we get cd^4/8D^3n.
P.mohan said:
9 years ago
Explain how to derive it?
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