A 150 N load is acting on a close coiled helical spring. The mean coil diameter has to be 12 times that of the wire diameter. If the maximum shear stress is not to exceed 150 N/mm², what is the diameter of the coil ?
- A57·64 mm
- B67·64 mmCorrect
- C77·64 mm
- D87·64 mm
Solution
Concept
Shear stress in a close-coiled helical spring (with direct shear correction)
Formula
τ = (8 W D / π d³) (1 + d / 2D)
Step-by-step solution
1. D = 12 d, so τ = (8 × 150 × 12 d / π d³)(1 + 1/24) = (14 400 / π d²)(1.0417). 2. Set τ = 150: d² = 14 400 × 1.0417 / (π × 150) = 31.83, d = 5.64 mm. 3. D = 12 × 5.64 = 67.7 mm ≈ 67.64 mm.
Explanation
The maximum shear stress in a spring includes the direct shear component, so the factor (1 + d/2D) is used. Without it D would be 66.3 mm, which is not an option.
Why the other options are wrong
57.64, 77.64 and 87.64 mm differ from the calculated coil diameter by more than 10 mm.
Shortcut
D = 12 d; solve τ = 150 for d, then D = 12 d.
Common mistake
Ignoring the direct shear correction (gives 66.3 mm).
Correct answer: B) 67·64 mm