Solution to Problem 344 | Helical Springs

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

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

$\tau_{max} = \dfrac{16(500)(4)}{\pi (1^3)} \left[ \dfrac{4(8) - 1}{4(8) - 4} + \dfrac{0.615}{8} \right]$

$\tau_{max} = 12\,060.3 \, \text{psi} = 12.1 \, \text{ ksi}$           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