Question 14.2: Allowable Load of a Helical Compression Spring A helical com......

The spring index is C=D / d=0.5 / 0.0625=8 .

a. Through the use of Equation (14.12) and Table 14.2, we have

S_{u s}=A d^b           (14.12)

S_u=A d^b=186\left(0.0625^{-0.163}\right)=292  ksi

Then, by Table 14.3, S_{y s}=0.4(292)=117  ksi .

b. The allowable load is obtained by applying Equation (14.6) as

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

P_{\text {all }}=\frac{\tau_{\text {all }} \pi d^2}{8 K_s C}

where

\tau_{ all }=\frac{S_{y s}}{n}=\frac{117}{1.5}=78  ksi

K_s=1+\frac{0.615}{8}=1.077        (from Equation (14.7))

K_s=1+\frac{0.61 .5}{C}        (14.7)

Hence,

P_{\text {all }}=\frac{\pi(78,000)(0.0625)^2}{8(1.077)(8)}=13.9  lb