Solution to Problem 343 | 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 = 1.5 kN = 1500 N; R = 90 mm
d = 20 mm; n = 20 turns
m = 2R/d = 2(90)/20 = 9
 

$\tau_{max} = \dfrac{16(1500)(90)}{\pi(20^3)} \left( \dfrac{4(9) - 1}{4(9) - 4} + \dfrac{0.615}{9} \right)$

$\tau_{max} = 99.87 \, \text{MPa}$           answer
 

$\delta = \dfrac{64PR^3n}{Gd^4} = \dfrac{64(1500)(90^3)(20)}{83\,000(20^4)}$

$\delta = 105.4 \, \text{mm}$           answer