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Question

A helical coil spring with wire diameter d and mean coil diameter D is subjected to axial load. A constant ratio of D and d has to be maintained, such that the extension of the spring is independent of D and d. What is the ratio?

The correct answer is \(\frac{{{D^3}}}{{{d^4}}}\)

Understanding Helical Spring Deflection

When a helical coil spring is subjected to an axial load, it undergoes deflection or extension. The amount of deflection ($\delta$) depends on various factors, including the applied load, the material properties, and the geometry of the spring.

Formula for Helical Spring Deflection

The standard formula for the deflection ($\delta$) of a close-coiled helical spring under an axial load (P) is given by:

$\delta = \frac{64 P n D^3}{G d^4}$

Where:

  • P is the axial load applied to the spring.
  • n is the number of active turns in the spring.
  • D is the mean coil diameter.
  • d is the wire diameter.
  • G is the shear modulus of the spring material.

Condition for Independent Extension

The question states that the extension (deflection $\delta$) of the spring must be independent of both the mean coil diameter (D) and the wire diameter (d). Looking at the deflection formula:

$\delta = \frac{64 P n}{G} \times \frac{D^3}{d^4}$

In this formula, P, n, and G are typically constant for a given spring and load scenario. The geometric properties that influence deflection are D and d, specifically through the term $\frac{D^3}{d^4}$.

For the deflection $\delta$ to be independent of D and d, the term $\frac{D^3}{d^4}$ must effectively behave as a constant that does not change when D or d are varied (as long as the specified ratio is maintained). This implies that the ratio $\frac{D^3}{d^4}$ itself must be a constant value.

If the ratio $\frac{D^3}{d^4} = \text{Constant}$, then the deflection formula becomes:

$\delta = \left(\frac{64 P n}{G}\right) \times \text{Constant}$

Since P, n, and G are constant, and we are maintaining the ratio $\frac{D^3}{d^4}$ as a constant, the deflection $\delta$ will then be independent of the individual values of D and d. Instead, it will only depend on the constant ratio maintained between them, the applied load, the number of coils, and the material properties.

Therefore, for the extension of the spring to be independent of D and d while maintaining a constant ratio between D and d, that constant ratio must be $\frac{D^3}{d^4}$.

Conclusion

The ratio that must be maintained constant for the extension of the helical spring to be independent of the mean coil diameter (D) and the wire diameter (d) is $\frac{D^3}{d^4}$.

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Important Questions from Springs

  1. Spring stiffness is defined as the

  2. When the spring of a watch is wound it possess _____.

  3. Two helical tensile spring of the same material and also having identical mean coil diameter and weight, have wire diameters d and \(\frac d2\). The ratio of their stiffness is

  4. A compression spring is made of a music wire of 2 mm diameter having a shear strength and shear modulus of 800 MPa and 80 GPa respectively. The mean coil diameter is 20 mm, the free length is 40 mm and the number of active coils is 10. If the mean coil diameter is reduced to 10 mm, the stiffness of the spring is approximately

  5. Two springs of stiffness 100 N/m each are connected in series and support a mass of 2 kg. The natural frequency of the system will be

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