Question 14.4: Design of a Helical Compression Spring for Cyclic Loading Pr...

See Figures 14-18, 14-19 and Table 14-11.

1    Find the number of cycles that the spring will see over its design life.

N_{\text {life }}=1000 \frac{ rev }{\min }\left\lgroup\frac{60  min }{ hr } \right\rgroup \left\lgroup\frac{2080  hr }{ shift – yr } \right\rgroup (10  yr )=1.2 E 9  \text { cycles }          (a)

This large a number requires that an endurance limit for infinite life be used.

2    Find the mean and alternating forces from equation 14.16a:

\begin{array}{l} F_a=\frac{F_{\max }-F_{\min }}{2} \\ F_m=\frac{F_{\max }+F_{\min }}{2} \end{array}        (14.16a)

\begin{array}{c} F_a=\frac{F_{\max }-F_{\min }}{2}=\frac{150-60}{2}=45  lb \\ F_m=\frac{F_{\max }+F_{\min }}{2}=\frac{150+60}{2}=105  lb \end{array}         (b)

3    Assume a 0.207-in wire diameter from the available sizes in Table 14-2 (p. 791). Assume a spring index of 9 and calculate the mean coil diameter D from equation 14.5 (p. 797).

C=\frac{D}{d}      (14.5)

D=C d=9(0.207)=1.863  \text { in }        (c)

4    Find the direct shear factor K_{s} and use it to calculate the stress \tau _{i} at the initial deflection (lowest defined force), and the mean stress \tau _{m}:

K_s=1+\frac{0.5}{C}=1+\frac{0.5}{9}=1.056      (d)

\tau_i=K_s \frac{8 F_i D}{\pi d^3}=1.056 \frac{8(60)(1.863)}{\pi(0.207)^3}=33875  psi       (e)

\tau_m=K_s \frac{8 F_m D}{\pi d^3}=1.056 \frac{8(105)(1.863)}{\pi(0.207)^3}=59281  psi       (f)

5    Find the Wahl factor K_{w} and use it to calculate the alternating shear stress \tau _{a} in the coil.

K_w=\frac{4 C-1}{4 C-4}+\frac{0.615}{C}=\frac{4(9)-1}{4(9)-4}+\frac{0.615}{9}=1.162       (g)

\tau_a=K_w \frac{8 F_a D}{\pi d^3}=1.162 \frac{8(45)(1.863)}{\pi(0.207)^3}=27970  psi          (h)

6    Find the ultimate tensile strength of this music-wire material from equation 14.3 and Table 14-4 (p. 792) and use it to find the ultimate shear strength from equation 14.4 and the torsional yield strength from Table 14-8 (p. 803), assuming that the set has been removed and using the low end of the recommended range.

S_{u t} \cong A d^b       (14.3)

S_{u s} \cong 0.67 S_{u t}          (14.4)

\begin{array}{l} S_{u t}=A d^b=184649(0.207)^{-0.1625}=238507  psi \\ S_{u s}=0.67 S_{u t}=159800  psi \end{array}       (i)

S_{y s}=0.60 S_{u t}=0.60(238507)=143104  psi          (j)

7    Find the wire endurance limit for peened springs in repeated loading from equation 14.13 (p. 804) and convert it to a fully reversed endurance strength with equation 14.18b (p. 814).

\begin{array}{l} S_{e w^{\prime}} \cong 45.0  kpsi (310  MPa ) \quad \text {for unpeened springs} \\ S_{e w^{\prime}} \cong 67.5  kpsi (465  MPa ) \quad \text {for peened springs} \end{array}        (14.13)

S_{e s}=0.5 \frac{S_{e w} S_{u s}}{S_{u s}-0.5 S_{e w}}        (14.18b)

S_{e w}=67500  psi       (k)

S_{e s}=0.5 \frac{S_{e w} S_{u s}}{S_{u s}-0.5 S_{e w}}=0.5 \frac{67500(159800)}{159800-0.5(67500)}=42787  psi      (l)

8    The safety factor is calculated from equation 14.18a (p. 814).

N_{f_s}=\frac{S_{e s}\left(S_{u s}-\tau_i\right)}{S_{e s}\left(\tau_m-\tau_i\right)+S_{u s} \tau_a}         (14.18a)

\begin{aligned} N_{f s} &=\frac{S_{e s}\left(S_{u s}-\tau_i\right)}{S_{e s}\left(\tau_m-\tau_i\right)+S_{u s} \tau_a} \\ &=\frac{42787(159800-33875)}{42787(59281-33875)+159800(27970)}=1.0 \end{aligned}          (m)

This is obviously not an acceptable design. To get some idea of what to change to improve it, the model was solved for a list of values of the spring index C from 4 to 14, keeping all other parameters as defined above. The resulting values of coil diameter, free length, spring weight, and torsional fatigue safety factor are plotted in Figure 14-18. Note that the wire diameter was held constant to develop the variation of parameters with spring index as shown in Figure 14-18. If another parameter such as mean coil diameter D were held constant instead, a different set of functions would result for the free length, weight, safety factor, etc.

The safety factor increases with decreasing spring index, so a reduction in our assumed value for C will improve the safety factor even with no change in wire diameter. Note, however, that the free length increases exponentially with decreasing spring index. If package size is limited, we may not want to decrease the spring index too much in order to avoid excessive spring length. The coil diameter increases linearly with the spring index for a constant wire diameter. Spring weight decreases slowly with increasing spring index.

If we decrease the spring index from 9 to 7, keeping all other parameters the same, we will obtain an acceptable design in this case with N_{f} = 1.3. Table 14-11 shows the results of this new calculation and of the calculations outlined below needed to complete the design. Figure 14-19 shows the modified-Goodman diagram for the final design. A summary of the changed values is

\begin{array}{llll} C=7 & D=1.45  in & K_w=1.21 & K_s=1.07 \\ \tau_i=26743  psi & \tau_a=22705  psi & \tau_m=46800  psi & N_{f_s}=1.3 \end{array}       (n)

9    The spring rate is defined from the two specified forces at their relative deflection.

k=\frac{\Delta F}{y}=\frac{150-60}{1.0}=90  lb / \text { in }         (o)

10    To get the defined spring rate, the number of active coils must satisfy equation 14.7 (p. 797):

k=\frac{F}{y}=\frac{d^4 G}{8 D^3 N_a}         (14.7)

k=\frac{d^4 G}{8 D^3 N_a} \quad \text { or } \quad N_a=\frac{d^4 G}{8 D^3 k}=\frac{(0.207)^4 11.5 E 6}{8(1.45)^3(90)}=9.64 \cong 9 \frac{3}{4}         (p)

Note that we round it to the nearest 1/4 coil, as the manufacturing tolerance cannot achieve better than that accuracy. This makes the spring rate k = 89 lb/in.

11    Assume squared and ground ends, making the total number of coils from Figure 14-9 (p. 796):

N_t=N_a+2=9.75+2=11.75       (q)

12    The shut height can now be determined.

L_s=d N_t=0.207(11.75)=2.43  \text { in }       (r)

13    The initial deflection to reach the smaller of the two specified loads is

y_{\text {initial }}=\frac{F_{\text {initial }}}{k}=\frac{60}{89}=0.674  \text { in }          (s)

14    Assume a clash allowance of 15% of the working deflection:

y_{\text {clash }}=0.15 y=0.15(1.0)=0.15  \text { in }       (t)

15    The free length (see Figure 14-8, p. 796) can now be found from

L_f=L_s+y_{\text {clash }}+y_{\text {working }}+y_{\text {initial }}=1.82+0.15+1.0+0.674=4.25  \text { in }       (u)

16    The deflection to the shut height is

y_{s h u t}=L_f-L_s=4.25-2.43=1.82  \text { in }        (v)

17    The force at this shut-height deflection is

F_{\text {shut }}=k y_{\text {shut }}=89(1.82)=162  lb        (w)

18    The shut-height stress and safety factor are

\tau_{\text {shut }}=K_s \frac{8 F_{\text {shut }} D}{\pi d^3}=1.07 \frac{8(164)(1.45)}{\pi(0.207)^3}=72875  psi           (x)

N_{s_{\text {shut }}}=\frac{S_{y s}}{\tau_{\text {shut }}}=\frac{143104  psi }{72875  psi }=2.0         (y)

which is acceptable.

19    To check for buckling, two ratios need to be calculated, L_{f} / D and y_{max} / L_{f}.

\begin{array}{c} \frac{L_f}{D}=\frac{4.25}{1.45}=2.93 \\ \frac{y_{\max }}{L_f}=\frac{y_{\text {initial }}+y_{\text {working }}}{L_f}=\frac{0.674+1.0}{4.25}=0.39 \end{array}          (z)

Take these two values to Figure 14-14 (p. 801) and find that their coordinates are safely within the zones that are stable against buckling for either end-condition case.

20    The weight of the spring’s active coils from equation 14.12b (p. 802) is

W_a=\frac{\pi^2 d^2 D N_a \gamma}{4}       (14.12b)

W_a=\frac{\pi^2 d^2 D N_a \gamma}{4}=\frac{\pi^2(0.207)^2(1.45)(9.75)(0.285)}{4}=0.426  lb            (aa)

21    The natural frequency of this spring is found from equation 14.12a (p. 802) and is

\omega_n=\pi \sqrt{\frac{k g}{W_a}}  rad / sec \quad f_n=\frac{1}{2} \sqrt{\frac{k g}{W_a}}  Hz         (14.12a)

f_n=\frac{1}{2} \sqrt{\frac{k g}{W}}=\frac{1}{2} \sqrt{\frac{89(386)}{0.426}}=142 Hz =8521 \frac{\text { cycles }}{\text { min }}          (ab)

The ratio between the natural frequency and the forcing frequency is

\frac{8520}{1000}=8.5         (ac)

which is sufficiently high.

22    The design specification for this A228 wire spring is

d=0.207  \text { in } \quad D_o=1.66  \text { in } \quad N_t=11.75, sq \& g \quad L_f=4.25      (ad)

23    Detailed results are shown in Table 14-11. The files EX14-04 are on the CD-ROM. There are also alternate approaches to the solution of this example shown in separate Mathcad and TKSolver files on the CD-ROM with a letter added to their name.