Civil Engineering - UPSC Civil Service Exam Questions - Discussion


In a close-coiled helical spring subjected to an axial load, other quantities remaining the same, if the wire diameter is doubled, then the stiffness of the spring when compared to the original one, will become

[A]. twice
[B]. four times
[C]. eight times
[D]. sixteen times

Answer: Option D


No answer description available for this question.

Bapuji said: (Jul 16, 2018)  
Please, anyone explain the answer.

Mnb said: (Mar 14, 2019)  
Stiffness = Gd^4/64R^3n.
2^4 = 16.

Shubham said: (Apr 5, 2020)  
It should be twice.

Ajay Sharma said: (May 17, 2021)  
K = p/(δ) in spring.
The formula is k=Gd^4/(64R^3n).
So, (2d)^4= 16d^4.

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