# Civil Engineering - Theory of Structures - Discussion

Discussion Forum : 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
25 N
30 N
35 N
40 N
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
7 comments Page 1 of 1.

Chulbul said:   7 years ago
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.
(1)

Aamir said:   7 years ago
It contains 8 coils, and 1 coil contains 7 clearances there total clearance =56=56mm.
(2)

Snehal Wankhede said:   8 years ago
For 1 clearance p = 5, 8 coil = 7 clearance,
So, p total = 35.
(2)

Akit pads said:   8 years ago
@Ashish.

By using this formula, I got an answer as 5N. Am I right?
(2)

Ashish said:   8 years ago
The formula is : (64 * p * (r^3) * n)/Nd4.
(1)

Achyut Paudel said:   9 years ago
Can anyone solve this question with explanation?

Rashmi said:   9 years ago
8 = [64*P*(40^3)*8]/(80000*(4^4)).

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