A round wire helical spring is loaded by the axial force. The maximum shear stress in the wire cross section will be at

Correct option is B

In a round wire helical spring subjected to an axial load, the maximum shear stress occurs at the inner surface of the wire's cross-section.

Stress Concentration: The curvature of the helical coil creates higher stress on the inner side of the wire.
Superposition: While torsional stress is highest on the outer fiber, the combined effect of torsional and direct shear stresses (modified by curvature) shifts the peak stress to the inner fiber.

$$\begin{aligned}&\text{The total shear stress is the sum of torsional and direct shear stresses. However, their distributions differ:} \\&\bullet\ \textbf{Torsional stress} \text{ peaks at the \textit{outer surface}.} \\&\bullet\ \textbf{Direct shear stress} \text{ is uniform but \textit{adds constructively at the inner surface} due to the spring's curvature.} \\[1em]&\textbf{Wahl Correction Factor } (K) \\[0.5em]&\text{For helical springs, the \textit{curvature of the wire} amplifies the stress at the \textit{inner surface}.} \\&\text{The Wahl factor accounts for this:} \\&K = \frac{4C - 1}{4C - 4} + \frac{0.615}{C} \\[1em]&\text{where } C = \frac{2R}{d} \text{ (spring index)}. \\[1em]&\text{The maximum shear stress then becomes:} \\&\tau_{\text{max}} = K \cdot \frac{8FD}{\pi d^3} \\[1em]&\text{where } D = 2R \text{ is the mean coil diameter.}\end{aligned}$$​​