Question 14.3: Design of a Hard-Drawn Wire Compression Spring A helical com......

The direct shear factor, from Equation (14.7), is K_s =1+(0.615/9)=1.068. The ultimate strength is estimated using Equation (14.12) and Table 14.2 as

K_s=1+\frac{0.61 .5}{C}        (14.7)

S_{u s}=A d^b         (14.12)

S_u=A d^b=1.51\left(10^9\right) d^{-0.201}

in which d is in millimeters. Expressing d in meters, the foregoing becomes

S_u=1.51\left(10^9\right)(1000)^{-0.201} d^{-0.201}=376.7\left(10^6\right) d^{-0.201}

The yield strength in shear, referring to Table 14.3, is then

S_u=0.42 S_u=158.2\left(10^6\right) d^{-0.201}        (a)

Substitution of the given numerical values into Equation (14.6) together with \tau_{ all } / n , the maximum design shear stress is expressed as

\tau_t=K_s \frac{8 P D}{\pi d^3}=K_s \frac{8 P C}{\pi d^2}          (14.6)

\begin{aligned} \tau_{\text {all }} & =\frac{8 n K_s C P_{\text {all }}}{\pi d^2} \\ & =\frac{8(1.8)(1.068 \times 9)(45)}{\pi d^2}=\frac{1982.6}{d^2} \end{aligned}       (b)

Finally, equating Equations (a) and (b) results in

158.2(10)^6 d^{-0.201}=1982.6 d^{-2}

from which

d=0.00188  m =1.88  mm

Thus, the mean coil diameter equals

D=C d=9(1.88)=16.92  mm

Comment: A standard 1.9 mm diameter hard-drawn wire should be used.