**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.

Step-by-Step

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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. |

Question: 14.8

Equation (14.43) may be rearranged into the form
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