Discussion :: Theory of Structures - Section 1 (Q.No.25)
| 25. | The stiffness of the close coil helical spring is |
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Answer: Option A Explanation: No answer description available for this question.
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| 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. |
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| 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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