Question 14.7: Spiral Torsion Spring: Design for Static Loading For a torsi......

By Equation (14.12) and Table 14.2,

S_{u s}=A d^b          (14.12)

S_u A d^b=2060\left(1.625^{-0.163}\right)=1903  MPa

From Equation (7.5b) and Table 14.3,

S_{y s}=0.577 S_y      (7.5b)

S_y=\frac{S_{y s}}{0.577}=0.4 \frac{1903}{0.577}=1319  MPa

Applying Equation (14.36) with C =25/1.625=15.38,

\begin{gathered} K_i=\frac{4 C^2-C-1}{4 C(C-1)} \\ K_o=\frac{4 C^2+C-1}{4 C(C+1)} \end{gathered}          (14.36)

K_i=\frac{4(15.38)^2-15.38-1}{4(15.38)(15.38-1)}=1.051

Through the use of Equation (14.39), we have

\sigma_i=\frac{32 P a}{\pi d^3} K_i     (round wire)      (14.39)

\begin{aligned} M & =P a=\frac{\pi d^3 S_y / n}{32 K_i}=\frac{\pi(1.625)^3(1319 / 1.5)}{32(1.051)} \\ & =352.5  N \cdot mm \end{aligned}

The geometric properties of the spring are L_{ w }=\pi D N_a=\pi(25)(350)=27,489 \quad mm \quad \text { and } \quad I=\pi (1.625)^4 / 64=0.342  mm ^4 . Equation (14.41) results in

\theta_{ rev }=\frac{1}{2 \pi} \theta_{ rad }=\frac{1}{2 \pi} \frac{M L_w}{E I}        (14.41)

\theta_{ rad }=\frac{M L_w}{E I}=\frac{352.5(27,489)}{\left(207 \times 10^3\right)(0.342)}=136.9  rad

Comment: The maximum moment winds the spring 136.9 / 2 \pi=21.8 turns.