Question

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

• A. \(\dfrac{8FD}{\pi d^3} + \dfrac{4F}{\pi d^2}\)
• B. \(\dfrac{8FD}{\pi d^3} + \dfrac{2F}{\pi d^2}\)
• C. \(\dfrac{16FD}{\pi d^3} + \dfrac{4F}{\pi d^2}\)
• D. \(\dfrac{32FD}{\pi d^3} + \dfrac{4F}{\pi d^2}\)

Hint:

Add the torsional shear stress from the twisting moment to the direct shear stress from the axial load.

Answer 1

The Correct Option is A

Step 1: Identify the two stress components in the wire.
A helical compression spring wire under axial load \(F\) carries a torsional shear stress from the twisting moment \(T = F \times D/2\), plus a direct transverse shear stress from the force \(F\) acting on the wire cross section.

Step 2: Write the torsional shear stress.
For a circular wire of diameter \(d\), the torsional shear stress is \(\tau_t = \dfrac{16T}{\pi d^3} = \dfrac{16 \times FD/2}{\pi d^3} = \dfrac{8FD}{\pi d^3}\).

Step 3: Write the direct shear stress.
The direct shear stress from the axial load spread over the wire cross section area \(\pi d^2/4\) is \(\tau_d = \dfrac{F}{\pi d^2/4} = \dfrac{4F}{\pi d^2}\).

Step 4: Add the two components.
Both stresses act on the same section and add on the inner side of the coil, so the maximum shear stress is \(\tau_{max} = \tau_t + \tau_d = \dfrac{8FD}{\pi d^3} + \dfrac{4F}{\pi d^2}\).

Final Answer:
This matches the torsional plus direct shear expression in option A. \[ \boxed{\tau_{max} = \dfrac{8FD}{\pi d^3} + \dfrac{4F}{\pi d^2}} \]