## Bending Stress

Made from a 0.125 inch diameter wire, a helical torsion spring has a mean spring diameter of 2 inches. If a 0.5 lb load is applied 3 inches away from the coil’s center, what is the spring’s bending stress in lb/in^2?

Expand Hint
For a helical torsion spring, the bending stress is:
$$\sigma=K_i(\frac{32Fr}{\pi d^3})$$$where $$K_i$$ is the correction factor, $$F$$ is the applied load, $$r$$ is the radius from the coil’s center to the load, and $$d$$ is the wire diameter. Hint 2 Correction factor: $$K_i=\frac{(4C^2-C-1)}{[4C(C-1)]}$$$
where $$C=D/d$$ , $$D$$ is the mean spring diameter, and $$d$$ is the wire diameter.
Correction factor:
$$K_i=\frac{(4C^2-C-1)}{[4C(C-1)]}$$$where $$C=D/d$$ , $$D$$ is the mean spring diameter, and $$d$$ is the wire diameter. First, let’s solve for $$C$$ : $$C=\frac{2in}{0.125in}=16$$$
Next, solve for the correction factor:
$$K_i=\frac{(4(16)^2-(16)-1)}{[4(16)(16-1)]}=\frac{(1,024-16-1)}{[64(15)]}=\frac{1,007}{960}=1.0489$$$For a helical torsion spring, the bending stress is: $$\sigma=K_i(\frac{32Fr}{\pi d^3})$$$
where $$K_i$$ is the correction factor, $$F$$ is the applied load, $$r$$ is the radius from the coil’s center to the load, and $$d$$ is the wire diameter.
$$\sigma=1.0489\times \frac{32(0.5lb)(3in)}{\pi (0.125in)^3}=1.0489 \times \frac{48lb}{0.00613in^2}$$$$$=1.0489 \times 7,826.75\frac{lb}{in^2}=8,210\:\frac{lb}{in^2}$$$
$$8,210\:\frac{lb}{in^2}$$\$