A 10 coils, 120 mm mean diameter closely coiled helical spring is made up of a 12 mm steel wire. If the spring is carrying an axial load of 150 N, what is the maximum shear stress ? (Take modulus of rigidity C = 8·16 × 10⁴ N/mm²)
- A16·52 N/mm²
- B20·52 N/mm²
- C26·52 N/mm²Correct
- D32·52 N/mm²
Solution
Concept
Maximum shear stress in a helical spring
Formula
τ = 8 W D / (π d³)
Step-by-step solution
1. τ = 8 × 150 × 120 / (π × 12³). 2. = 144 000 / 5 428.7 = 26.52 N/mm².
Explanation
The torsional shear stress in the wire of a close-coiled spring is τ = 8WD/(πd³). The number of coils and the modulus of rigidity are not needed for the stress.
Why the other options are wrong
The other values do not satisfy 8WD/(πd³) = 26.52.
Shortcut
τ = 8WD/(πd³).
Common mistake
Using D = 12 instead of 120 mm.
Correct answer: C) 26·52 N/mm²