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The Wahl's stress factor (K) for springs is given by
(where $C = Spring index = \frac { Mean dia. of spring (D) } { dia. of wire (d)}$

$\frac { 2C + 1 } { 2C- 1} + \frac {0.615}{C}$

$\frac { 2C - 1 } { 2C + 1} + \frac {0.615}{C}$

$\frac { 4C + 1 } { 4C- 4} + \frac {0.615}{C}$

$\frac { 4C - 1 } { 4C - 4} + \frac {0.615}{C}$

Answer : Option D

Explanation :

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