Solution to Problem 346 | Helical Springs

$\delta = \dfrac{64PR^3n}{Gd^4}$
 

Where
δ = 100 mm; R = 100 mm
d = 20 mm; n = 24 turns
G = 42 000 MPa
 

$100 = \dfrac{64P(100^3)24}{42\,000(20^4)}$

$P = 437.5 \, \text{N}$
 

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

Where
m = 2R/d = 2(100)/20 = 10
 

$\tau_{max} = \dfrac{16(437.5)(100)}{\pi (20^3)} \left[ \dfrac{4(10) - 1}{4(10) - 4} + \dfrac{0.615}{10} \right]$

$\tau_{max} = 31.89 \, \text{ MPa}$           answer