Question 14.5: Design of a Helical Extension Spring for Cyclic Loading Prob...

See Figure 14-24 (p. 828) and Table 14-14 (pp. 829-830).

1    Assume an 0.177 in trial wire diameter from the available sizes in Table 14-2 (p. 791). Assume a spring index of C = 9 and use it to calculate the mean coil diameter D from equation 14.5 (p. 797).

C=\frac{D}{d}     (14.5)

D=C d=9(0.177)=1.59  \text { in }        (a)

2    Use the assumed value of C to find an appropriate value of initial coil stress \tau _{i} from equations 14.21 (p. 821):

k=\frac{F-F_i}{y}=\frac{d^4 G}{8 D^3 N_a}       (14.21)

\begin{aligned} \tau_{i_1} & \cong-4.231 C^3+181.5 C^2-3387 C+28640 \\ &=-4.231(9)^3+181.5(9)^2-3387(9)+28640=9774  \text { psi } \end{aligned}          (b)

\begin{aligned} \tau_{i_2} & \cong-2.987 C^3+139.7 C^2-3427 C+38404 \\ &=-2.987(9)^3+139.7(9)^2-3427(9)+38404=16699  psi \end{aligned}        (c)

\tau_i \cong \frac{\tau_{i_1}+\tau_{i_2}}{2}=\frac{9774+16699}{2}=13237  psi        (d)

3    Find the direct shear factor:

K_s=1+\frac{0.5}{C}=1+\frac{0.5}{9}=1.06          (e)

4    Substitute K_{s} from step 3 and the value of \tau _{i} from equation (d) of step 2 for \tau _{max} in equation 14.8b (p. 798) to find the corresponding initial coil-tension force F_{i}:

\begin{aligned} \tau_{\max } &=\frac{8 F C}{\pi d^2}+\frac{4 F}{\pi d^2}=\frac{8 F C+4 F}{\pi d^2} \\ &=\frac{8 F C}{\pi d^2}\left\lgroup1+\frac{1}{2 C}\right\rgroup =\frac{8 F D}{\pi d^3}\left\lgroup1+\frac{0.5}{C}\right\rgroup \\ \tau_{\max } &=K_s \frac{8 F D}{\pi d^3} \quad \quad \text { where } \quad K_s=\left\lgroup1+\frac{0.5}{C}\right\rgroup \end{aligned}        (14.8b)

F_i=\frac{\pi d^3 \tau_i}{8 K_s D}=\frac{\pi(0.177)^3(13237)}{8(1.06)(1.59)}=17.1  lb         (f)

Check that this force is less than the required minimum applied force F_{min}, which in this case, it is. Any applied force smaller than F_{i} will not deflect the spring.

5    Find the mean and alternating forces from equation 14.16a (p. 812):

\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{85-50}{2}=17.5  lb \\ F_m=\frac{F_{\max }+F_{\min }}{2}=\frac{85+50}{2}=67.5  lb \end{array}        (g)

6    Use the direct shear factor K_{s} and previously assumed values to find the mean stress* \tau _{m}:

\tau_m=K_s \frac{8 F_m D}{\pi d^3}=1.06 \frac{8(67.5)(1.59)}{\pi(0.177)^3}=52122  psi        (h)

7    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}=1.16         (i)

\tau_a=K_w \frac{8 F_a D}{\pi d^3}=1.16 \frac{8(17.5)(1.59)}{\pi(0.177)^3}=14877  psi        (j)

8    Find the ultimate tensile strength of this music-wire material from equation 14.3 and Table 14-4 (p. 792). Use it to find the ultimate shear strength from equation 14.4 and the torsional yield strength for the coil body from Table 14-12, assuming no set removal.

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(0.177)^{-0.1625}=244633  psi \\ S_{u s}=0.667 S_{u t}=163918  psi \end{array}        (k)

S_{y s}=0.45 S_{u t}=0.45(244633)=110094  psi     (l)

9    Find the wire endurance limit for unpeened springs from equation 14.14 (p. 805) and convert it to a fully reversed endurance strength with equation 14.18b (p. 814).

S_{m s} \cong 0.9 S_{u s} \cong 0.9\left(0.67 S_{u t}\right) \cong 0.6 S_{u t}      (14.14)

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}=45000  psi       (m)

S_{e s}=0.5 \frac{S_{e w} S_{u s}}{S_{u s}-0.5 S_{e w}}=0.5 \frac{45000(163918)}{163918-0.5(45000)}=26080  psi          (n)

10    The fatigue safety factor for the coils in torsion 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{array}{c} C_1=\frac{2 R_1}{d}=\frac{2 D}{2 d}=C=9 \\ K_b=\frac{4 C_1^2-C_1-1}{4 C_1\left(C_1-1\right)}=\frac{4(9)^2-(9)-1}{4(9)(9-1)}=1.09 \end{array}         (o)

Note that the minimum stress due to force F_{min} is used in this calculation, not the coil-winding stress from equation (d).

11    The stresses in the end hooks also need to be determined. The bending stresses in the hook are found from equation 14.24 (p. 823):

\sigma_A=K_b \frac{16 D F}{\pi d^3}+\frac{4 F}{\pi d^2}      (14.24a)

\text {were} \quad  K_b=\frac{4 C_1^2-C_1-1}{4 C_1\left(C_1-1\right)}        (14.24b)

\text {and} \quad  C_1=\frac{2 R_1}{d}        (14.24c)

\begin{array}{c} C_1=\frac{2 R_1}{d}=\frac{2 D}{2 d}=C=9 \\ K_b=\frac{4 C_1^2-C_1-1}{4 C_1\left(C_1-1\right)}=\frac{4(9)^2-(9)-1}{4(9)(9-1)}=1.09 \end{array}         (p)

\begin{array}{l} \sigma_a=K_b \frac{16 D F_a}{\pi d^3}+\frac{4 F_a}{\pi d^2}=1.09 \frac{16(1.59)(17.5)}{\pi(0.177)^3}+\frac{4(17.5)}{\pi(0.177)^2}=28626  psi \\ \sigma_m=K_b \frac{16 D F_m}{\pi d^3}+\frac{4 F_m}{\pi d^2}=1.09 \frac{16(1.59)(67.5)}{\pi(0.177)^3}+\frac{4(67.5)}{\pi(0.177)^2}=110416  psi \end{array}          (q)

\sigma_{\min }=K_b \frac{16 D F_{\min }}{\pi d^3}+\frac{4 F_{\min }}{\pi d^2}=1.09 \frac{16(1.59)(50)}{\pi(0.177)^3}+\frac{4(50)}{\pi(0.177)^2}=81790  psi         (r)

12    Convert the torsional endurance strength to a tensile endurance strength with equation 14.4 (p. 793) and use it and the ultimate tensile strength from step 8 in equation 14.18 (p. 814) to find a fatigue safety factor for the hook in bending:

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

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)

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

\begin{aligned} S_e &=\frac{S_{e s}}{0.67}=\frac{26080}{0.67}=38925  psi \\ N_{f_b} &=\frac{S_e\left(S_{u t}-\sigma_{\min }\right)}{S_e\left(\sigma_{\text {mean }}-\sigma_{\min }\right)+S_{u t} \sigma_{\text {alt }}} \\ &=\frac{38925(244633-81790)}{38925(110416-81790)+244633(28626)}=0.78 \end{aligned}           (s)

13    The torsional stresses in the hook are found from equation 14.25 (p. 823) using an assumed value of C_{2} = 5.

\tau_B=K_{w_2} \frac{8 D F}{\pi d^3}       (14.25a)

\text {where} \quad  K_{W_2}=\frac{4 C_2-1}{4 C_2-4}        (14.25b)

\text {and}  \quad  C_2=\frac{2 R_2}{d}         (14.25c)

\begin{array}{l} R_2=\frac{C_2 d}{2}=\frac{5(0.177)}{2}=0.44  \text { in } \\ K_{w_2}=\frac{4 C_2-1}{4 C_2-4}=\frac{4(5)-1}{4(5)-4}=1.2 \end{array}         (t)

\begin{aligned} \tau_{B_a} &=K_{w_2} \frac{8 D F_a}{\pi d^3}=1.19 \frac{8(1.59) 17.5}{\pi(0.177)^3}=15202  psi \\ \tau_{B_m} &=K_{w_2} \frac{8 D F_m}{\pi d^3}=1.19 \frac{8(1.59) 67.5}{\pi(0.177)^3}=58637  psi \\ \tau_{B_{\min }} &=K_{w_2} \frac{8 D F_{\min }}{\pi d^3}=1.19 \frac{8(1.59) 50}{\pi(0.177)^3}=43435  psi \end{aligned}         (u)

14    The hook’s fatigue safety factor in torsion 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)

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

\begin{aligned} N_{f_s} &=\frac{S_{e s}\left(S_{u s}-\tau_{\min }\right)}{S_{e s}\left(\tau_m-\tau_{\min }\right)+S_{u s} \tau_a} \\ &=\frac{26080(163918-43435)}{26080(58637-43435)+163918(15202)}=1.1 \end{aligned}         (v)

15    One of these safety factors is less than 1, making this an unacceptable 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 from 4 to 14, keeping all other parameters as defined above. The resulting values of coil diameter, free length, spring weight, and fatigue safety factor are plotted in Figure 14-24.

The safety factors decrease with increasing spring index, so a reduction in our assumed value for C will improve the design even with no change in wire diameter. Note, however, that the spring free length shows a minimum value at a spring index of about 7.5. The coil diameter increases linearly with the spring index for a constant wire diameter. Spring weight decreases with increasing spring index.

If we decrease the spring index from 9 to 7.5, and increase the wire diameter one size to 0.192 in, keeping all other parameters the same, we will obtain an acceptable design in this case with the smallest N_{f} = 1.2 for the hook in bending.

16    Table 14-14 shows the complete results for this new design. A summary of the new design is

\begin{array}{l} C=7.5 \quad D=1.44  \text { in } \quad K_w=1.20 \quad K_s=1.07\\ \tau_i=15481  psi \quad \tau_{\min }=27631  psi \quad \tau_a=10856  psi \quad \tau_m=37302  psi\\ N_{s_{\text {coil }}}=2.3 \quad N_{f_{s_{c o i l}}}=1.8 \quad N_{f_{s_{\text {hook }}}}=1.7 \quad N_{f_{b_{\text {hook }}}}=1.2 \end{array}          (w)

The spring design can now be completed based on the new wire diameter and spring index from step 15.

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

k=\frac{\Delta F}{y}=\frac{85-50}{0.5}=70  lb / \text { in }       (x)

18    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{(0.192)^4 11.5 E 6}{8(1.44)^3(70)}=9.35 \cong 9 \frac{1}{4}        (y)

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 = 70.7 lb/in.

19    The total number of coils in the body and the body length are

\begin{array}{l} N_t=N_a+1=9.25+1=10.25 \\ L_b=N_t d=10.25(0.192)=1.97  in \end{array}         (z)

20    The free length can now be determined. The length of a standard hook is equal to the coil inside diameter:

L_f=L_b+2 L_{\text {hook }}=1.97+2(1.25)=4.46  \text { in }         (aa)

21    The deflection to reach the larger of the two specified loads is

y_{\max }=\frac{F_{\text {max }}-F_{\text {initial }}}{k}=\frac{85-28}{70.7}=0.81  \text { in }          (ab)

22    The natural frequency of this spring is found from equation 14.26 (p. 823) and is

f_n=\frac{2}{\pi N_a} \frac{d}{D^2} \sqrt{\frac{G g}{32 \gamma}} \quad Hz       (14.26)

f_n=\frac{2}{\pi N_a} \frac{d}{D^2} \sqrt{\frac{G g}{32 \gamma}}=\frac{2(0.192)}{\pi(9.25)(1.44)^2} \sqrt{\frac{11.5 E 6(386)}{32(0.285)}}=140.6  Hz =8436  rpm         (ac)

The ratio between the natural frequency and the forcing frequency is

\frac{8440}{500}=16.9        (ad)

which is sufficiently high.

23    The design specification for this A228 wire spring is

d=0.192  \text { in } \quad O D=1.63  \text { in } \quad N_t=10.25 \quad L_f=4.46         (ae)

24    The results are shown in Table 14-14. The files EX14-05 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.

* The direct shear factor K_{s} is used rather than the Wahl factor for the mean stress calculation because the stress concentration factor is considered to be 1.0 for mean stress.