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
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.
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.
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?
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)).
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers