Problem 345

A helical spring is fabricated by wrapping wire 3/4 in. in diameter around a forming cylinder 8 in. in diameter. Compute the number of turns required to permit an elongation of 4 in. without exceeding a shearing stress of 18 ksi. Use Eq. (3-9) and G = 12 × 10⁶ psi.

Solution 345

\tau_{max} = \dfrac{16PR}{\pi d^3} \left( 1 + \dfrac{d}{4R} \right) \, \to \,\,       Equation (3-9)

18\,000 = \dfrac{16P(4)}{\pi (3/4)^3} \left[ 1 + \dfrac{3/4}{4(4)} \right]

P = 356.07 \, \text{lb}

\delta = \dfrac{64PR^3n}{Gd^4}

4 = \dfrac{64(356.07)(4^3)n}{(12 \times 10^6)(3/4)^4}

n = 10.41 \, \text{say 10 turns} \,\,            answer