A closely coiled helical compression spring of mean coil diameter D and wire diameter d is loaded by axial force F, the maximum shear stress developed in the wire

The correct option is:
$$\frac{8FD}{\pi d^3}+\frac{4F}{\pi d^2}$$

Step-by-Step Derivation and Explanation:

When a closely coiled helical compression spring is subjected to an axial load, the wire of the spring experiences two primary types of shear stresses:
1. Torsional shear stress due to the twisting moment.
2. Direct shear stress due to the axial load acting on the cross-section of the wire.

Let us analyze these two stresses individually:

1. Torsional Shear Stress (τ₁)
The axial force F acts at the mean radius of the spring coil. This creates a twisting torque T on the wire. The torque is given by:
$$T=F\times \frac{D}{2}$$
where D is the mean coil diameter. For a circular wire section of diameter d, the polar section modulus Z_p is:
$$Z_p=\frac{\pi d^3}{16}$$
The torsional shear stress developed in the wire is:
$${\tau }_1=\frac{T}{Z_p}=\frac{F\times \frac{D}{2}}{\frac{\pi d^3}{16}}=\frac{8FD}{\pi d^3}$$

2. Direct Shear Stress (τ₂)
The axial force F acts directly on the cross-section of the wire. The cross-sectional area of the wire of diameter d is:
$$A=\frac{\pi d^2}{4}$$
The direct shear stress is therefore calculated as:
$${\tau }_2=\frac{F}{A}=\frac{F}{\frac{\pi d^2}{4}}=\frac{4F}{\pi d^2}$$

3. Combined Maximum Shear Stress (τ_max)
The maximum shear stress developed in the wire is the sum of the torsional shear stress and the direct shear stress:
$${\tau }_{\max }={\tau }_1+{\tau }_2$$
Substituting the derived expressions for τ₁ and τ₂, we get:
$${\tau }_{\max }=\frac{8FD}{\pi d^3}+\frac{4F}{\pi d^2}$$