# 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/mm^{2}), 4 mm in diameter. The clearance between the coils being 1 mm when unloaded, the minimum compressive load to remove the clearance, isDiscussion:

7 comments Page 1 of 1.
Chulbul said:
6 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.

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:
6 years ago

It contains 8 coils, and 1 coil contains 7 clearances there total clearance =56=56mm.

(1)

Snehal Wankhede said:
7 years ago

For 1 clearance p = 5, 8 coil = 7 clearance,

So, p total = 35.

So, p total = 35.

(1)

Akit pads said:
7 years ago

@Ashish.

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

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

(1)

Ashish said:
8 years ago

The formula is : (64 * p * (r^3) * n)/Nd4.

(1)

Achyut Paudel said:
8 years ago

Can anyone solve this question with explanation?

Rashmi said:
8 years ago

8 = [64*P*(40^3)*8]/(80000*(4^4)).

Post your comments here:

Your comments will be displayed after verification.

Quick links

Quantitative Aptitude

Verbal (English)

Reasoning

Programming

Interview

Placement Papers

© IndiaBIX™ Technologies