Question

In a closely coiled helical spring of circular wire, if d = diameter of spring wire; D = Mean diameter of spring coil; n = no. of active coils; C = spring index (D/d); G = Modulus of rigidity of the spring wire material, then the stiffness of the spring is:

• A. $\frac{G d}{8 C^3 n}$
• B. $\frac{G D}{8 C^2 n}$
• C. $\frac{8 C^3 n}{G D}$
• D. $\frac{8 C^3 n}{G d^2}$

Hint:

To remember the formula, look at the units. Rigidity ($G$) and diameter ($d$) are on top because they make the spring stronger/stiffer. The index ($C$) and number of coils ($n$) are on the bottom because adding more "springiness" or "width" makes it easier to compress.

Answer 1

The Correct Option is A

Stiffness, also known as the spring constant ($k$), is defined as the load required per unit deflection of the spring. For a closely coiled helical spring subjected to an axial load, the stiffness is derived from the torsion formula. 1. Basic Stiffness Formula: The standard formula for the stiffness of a helical spring is: $$k = \frac{G d^4}{8 D^3 n}$$

2. Incorporating the Spring Index (C): The spring index $C$ is defined as the ratio of the mean diameter of the coil to the diameter of the wire: $$C = \frac{D}{d} \implies D = C \cdot d$$

3. Derivation: Substituting $D = C \cdot d$ into the stiffness formula: $$k = \frac{G d^4}{8 (C d)^3 n}$$ $$k = \frac{G d^4}{8 C^3 d^3 n}$$ Simplifying by dividing both the numerator and denominator by $d^3$: $$k = \frac{G d}{8 C^3 n}$$

4. Conclusion: The stiffness of a spring is directly proportional to the modulus of rigidity and the wire diameter, but inversely proportional to the cube of the spring index and the number of active coils. Increasing the number of coils or the coil diameter makes the spring "softer" (less stiff).