# Civil Engineering - Theory of Structures - Discussion

### Discussion :: Theory of Structures - Section 1 (Q.No.40)

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

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.