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

A compression spring wire does not carry pure torsion alone. The axial force also produces a direct shear stress that adds to the torsional shear stress, and the exact form of that direct term is what separates the four choices.

1. $\dfrac{8FD}{\pi d^3} + \dfrac{4F}{\pi d^2}$: correct. The torque $FD/2$ on a circular section of diameter $d$ gives torsional stress $16T/(\pi d^3) = 8FD/(\pi d^3)$, and the direct shear stress from area $\pi d^2/4$ gives $4F/(\pi d^2)$. Adding both gives this expression.
2. $\dfrac{8FD}{\pi d^3} + \dfrac{2F}{\pi d^2}$: wrong, the direct shear term is understated here, the correct area based term is $4F/(\pi d^2)$, not half of it.
3. $\dfrac{16FD}{\pi d^3} + \dfrac{4F}{\pi d^2}$: wrong, this doubles the torsional term by skipping the moment arm of $D/2$, giving $16FD$ instead of $8FD$.
4. $\dfrac{32FD}{\pi d^3} + \dfrac{4F}{\pi d^2}$: wrong, this scales the torsional term far past the correct $8FD/(\pi d^3)$ value.

So the maximum shear stress in the wire is $\dfrac{8FD}{\pi d^3} + \dfrac{4F}{\pi d^2}$, option A.