Theory of Structures - Engineering

Q1:

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

ANS:C - 35 N

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.