Question 14.5: Helical Compression Spring: Design for Cyclic Loading A heli......

The mean and alternating loads are

P_m=\frac{1}{2}(600+300)=450  N , \quad P_a=\frac{1}{2}(600-300)=150  N

Equations (14.7) and (14.9) give

K_s=1+\frac{0.61 .5}{C}      (14.7)

K_w=\frac{4 C-1}{4 C-4}+\frac{0.615}{C}         (14.9)

K_s=1+\frac{0.615}{7}=1.088, \quad K_w=\frac{28-1}{28-4}+\frac{0.615}{7}=1.213

So we have, using Equations (14.21) and (14.22), \tau_a / \tau_m=K_w P_a / K_s P_m=0.372 .

\tau_m=K_s \frac{8 P_m C}{\pi d^2}       (14.21)

\tau_a=K_w \frac{8 P_a C}{\pi d^2}       (14.22)

a. Tentatively select a 6 mm wire diameter. Then from Equation (14.12) and Table 14.2, we have

S_{u s}=A d^b      (14.12)

S_u=A d^b=1790\left(6^{-0.155}\right)=1356  MPa

By Equation (7.5) and Table 14.3, S_{u s}=0.67(1356)=908.5  MPa \text { and } S_{e s}^{\prime}=0.2(1356)=271  MPa . Substitution of the numerical values into Equation (14.24) results in

   \begin{aligned} \text {Also Steels} \quad & S_{u s}=0.67 S_u   \quad (7.5a)\\ & S_{y s}=0.577 S_y   \quad (7.5b)\end{aligned}

\tau_a=\frac{S_{u s} / n}{\frac{\left(\tau_a / \tau_m\right)\left(2 S_{u s}-S_{e s}^{\prime}\right)}{S_{e s}^{\prime}}+1}        (14.24)

\tau_m=\frac{908.5 / 1.3}{\frac{(0.372)(2 \times 908.5-271)}{271}+1}=224  MPa

Applying Equation (14.26),

d^3=K_s \frac{8 P_m D}{\pi \tau_m} \quad \text { or } \quad d^2=K_s \frac{8 P_m C}{\pi \tau_m}       (14.26)

d^2=K_s \frac{8 P_m C}{\pi \tau_m}=1.088 \frac{8(450)(7)}{\pi\left(224 \times 10^6\right)}, \quad d=6.24\left(10^{-3}\right)  m

Hence, D =7(6.24)=43.68 mm. Inasmuch as S_u=1790\left(6.24^{-0155}\right)=1348<1356 MPa , d=6.24  mm is satisfactory

b. From Figure 14.7(d), h_s=\left(N_a+2\right) d=74.88 mm . Using Equation (14.11),

k=\frac{P}{\delta}=\frac{G d^4}{8 D^3 N_s}=\frac{d G}{8 C^3 N_a}         (14.11)

k=\frac{d G}{8 C^3 N_a}=\frac{(6.24)(79,000)}{8(7)^3(10)}=17.97  N / mm

With a 20% clash allowance,

\delta_s=1.2 \frac{P_{\max }}{k}=1.2(33.39)=40.07  mm

Thus,

h_f=74.88+40.07=115  mm

c. Through the use of Equation (14.29),

f_n=\frac{356,620 d}{D^2 N_a} Hz      (14.29)

\begin{aligned} f_n & =\frac{356,620 d}{D^2 N_a}=\frac{356,620(6.24)}{(43.68)^2(10)} \\ & =116.6 cps =6996  cpm \end{aligned}

Comment: If this corresponds to operating speeds (for equipment mounted on this spring), it may be necessary to redesign the spring.

d. Check for the buckling for extreme case of deflection \left(\delta=\delta_s\right):

\frac{\delta_s}{h_f}=\frac{40.07}{115}=0.35, \quad \frac{h_f}{D}=\frac{115}{43.68}=2.63

Since (2.63, 0.35) is inside of the stable region of curve A in Figure 14.10, the spring will not buckle.