Civil Engineering - Theory of Structures - Discussion

40. 

A spring of mean radius 40 mm contains 8 action coils of steel (N = 80000 N/mm2), 4 mm in diameter. The clearance between the coils being 1 mm when unloaded, the minimum compressive load to remove the clearance, is

[A]. 25 N
[B]. 30 N
[C]. 35 N
[D]. 40 N

Answer: Option C

Explanation:

No answer description available for this question.

Rashmi said: (Aug 4, 2015)  
8 = [64*P*(40^3)*8]/(80000*(4^4)).

Achyut Paudel said: (Nov 5, 2015)  
Can anyone solve this question with explanation?

Ashish said: (May 30, 2016)  
The formula is : (64 * p * (r^3) * n)/Nd4.

Akit Pads said: (Sep 9, 2016)  
@Ashish.

By using this formula, I got an answer as 5N. Am I right?

Snehal Wankhede said: (Sep 30, 2016)  
For 1 clearance p = 5, 8 coil = 7 clearance,
So, p total = 35.

Aamir said: (Oct 24, 2017)  
It contains 8 coils, and 1 coil contains 7 clearances there total clearance =56=56mm.

Chulbul said: (Jan 27, 2018)  
The question is asked to find the load. Not distance!

p= (64W* R^3)/(C*d^4).
We find out W, load for 1 clearance, between 2 coils.
Between 2 coil, 1 clearance, load= 5N.
Between 8 coil, 7 clearance, load= 5x7=35N.

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