If the wire diameter of a compressive helical spring is increased by 2%, the change in spring stiffness (in %) is _____ (correct to two decimal places).Correct answer is '8.00 to 8.50'. Can you explain this answer?

Question

Answer

Understanding Spring Stiffness
The stiffness of a helical spring is influenced by various factors, including the wire diameter. The formula for spring stiffness (k) is given by:

k = (G * d^4) / (8 * D^3 * n)
Where:
- G = Modulus of rigidity
- d = Wire diameter
- D = Mean coil diameter
- n = Number of active coils

Impact of Wire Diameter Increase
When the wire diameter (d) is increased by 2%, the new wire diameter becomes:

New d = d * (1 + 0.02) = 1.02d
To find the change in spring stiffness:
1. **Calculate the new stiffness (k')**:
k' = (G * (1.02d)^4) / (8 * D^3 * n)
2. **Express stiffness in terms of old stiffness (k)**:
k' = (G * d^4 * (1.02^4)) / (8 * D^3 * n)
k' = k * (1.02^4)
3. **Calculate the percentage increase in stiffness**:
- First, compute 1.02^4:
- 1.02^4 ≈ 1.08243216 (approximately 8.24%)
- Therefore, the change in spring stiffness is:
Percentage change = (k' - k) / k * 100
- This gives approximately 8.24%.

Conclusion
Thus, increasing the wire diameter by 2% results in a spring stiffness change of about 8.00 to 8.50%, confirming the correct answer range.

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