Problem 344

Determine the maximum shearing stress and elongation in a bronze helical spring composed of 20 turns of 1.0-in.-diameter wire on a mean radius of 4 in. when the spring is supporting a load of 500 lb. Use Eq. (3-10) and G = 6 × 10⁶ psi.

Solution 344

\tau_{max} = \dfrac{16PR}{\pi d^3} \left( \dfrac{4m - 1}{4m - 4} + \dfrac{0.615}{m} \right) \, \to \,\, Equation (3-10)

Where
P = 500 lb; R = 4 in
d = 1 in; n = 20 turns
m = 2R/d = 2(4)/1 = 8

\tau_{max} = \dfrac{16(500)(4)}{\pi (1^3)} \left[ \dfrac{4(8) - 1}{4(9) - 4} + \dfrac{0.615}{8} \right]
\tau_{max} = 12\,060.3 \, \text{psi} = 12.1 \, \text{ksi} \,\, answer

\delta = \dfrac{64PR^3n}{Gd^4} = \dfrac{64(500)(4^3)(20)}{(6 \times 10^6)(1^4)}
\delta = 6.83 \, \text{in} \,\, answer