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

Helical Compression Spring: Design for Cyclic Loading

A helical compression spring for a cam follower is subjected to the load that varies between $P_{min}$ and $P_{max}$. Apply the Goodman criterion to determine:

a. The wire diameter.

b. The free height.

c. The surge frequency.

d. Whether the spring will buckle in service.

Given: $P_{min}$=300 N, $P_{max}$=600 N.

Design Decisions: We use a chrome-vanadium ASTM A232 wire of $G$=79 GPa; $r_c$=20%, $N_a$=10, and  C [/;latex]=7. Both ends of the spring are squared and ground. A safety factor of 1.3 is used due to uncertainty about the load.

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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.

 TABLE 14.2 Coefficients and Exponents for Equation (14.12) $A$ Material ASTM No. $b$ MPa ksi Hard-drawn wire A227 −0.201 1510 237 Music wire A228 −0.163 2060 186 Oil-tempered wire A229 −0 193 1610 146 Chrome-vanadium wire A232 −0.155 1790 173 Chrome-silicon wire A401 −0 091 1960 218 Source: Associated Spring-Barnes Group, Design Handbook, Associated Spring-Barnes Group, Bristol, CN, 1987.

 TABLE 14.3 Approximate Strength Ratios of Some Common Spring Materials Material $S_{y s} / S_u$ $S_{e s}^{\prime} / S_u$ Hard-drawn wire 0.42 0.21 Music wire 0.40 0.23 Oil-tempered wire 0.45 0.22 Chrome-vanadium wire 0.52 0.20 Chrome-silicon wire 0.52 0.20 Source: Associated Spring-Barnes Group, Design Handbook, Associated Spring-Barnes Group, Bristol, CN, 1987. Notes: $S_{y s}$, yield strength in shear; $S_u,$ ultimate strength in tension; $S_{e s}^{\prime}$, endurance limit (or strength) in shear.

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