Design of a Helical Extension Spring for Cyclic Loading Problem Design an extension spring for a dynamic load over a given deflection. Given The spring must give a minimum force of 50 lb and a maximum force of 85 lb over a dynamic deflection of 0.50 in. The forcing frequency is 500 rpm. An infinite

Question

Design of a Helical Extension Spring for Cyclic Loading

Problem Design an extension spring for a dynamic load over a given deflection.

Given The spring must give a minimum force of 50 lb and a maximum force of 85 lb over a dynamic deflection of 0.50 in. The forcing frequency is 500 rpm. An infinite life is desired.

Assumptions Standard hooks will be used at each end. Music wire (ASTM A228) will be used since the loads are dynamic. Setting and peening cannot be used to obtain a higher endurance strength in an extension spring.

Accepted Answer

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 [latex]C[/latex] = 9 and use it to calculate the mean coil diameter [latex]D[/latex] from equation 14.5 (p. 797).

[latex] C=\frac{D}{d} [/latex] (14.5)

[latex] D=C d=9(0.177)=1.59 ext { in } [/latex] (a)

2 Use the assumed value of [latex]C[/latex] to find an appropriate value of initial coil stress [latex] au _{i}[/latex] from equations 14.21 (p. 821):

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

[latex] \begin{aligned} au_{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 ext { psi } \end{aligned} [/latex] (b)

[latex] \begin{aligned} au_{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} [/latex] (c)

[latex] au_i \cong \frac{ au_{i_1}+ au_{i_2}}{2}=\frac{9774+16699}{2}=13237 psi [/latex] (d)

3 Find the direct shear factor:

[latex] K_s=1+\frac{0.5}{C}=1+\frac{0.5}{9}=1.06 [/latex] (e)

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

[latex] \begin{aligned} au_{\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} ight group =\frac{8 F D}{\pi d^3}\left\lgroup1+\frac{0.5}{C} ight group \ au_{\max } &=K_s \frac{8 F D}{\pi d^3} \quad \quad ext { where } \quad K_s=\left\lgroup1+\frac{0.5}{C} ight group \end{aligned} [/latex] (14.8b)

[latex] F_i=\frac{\pi d^3 au_i}{8 K_s D}=\frac{\pi(0.177)^3(13237)}{8(1.06)(1.59)}=17.1 lb [/latex] (f)

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

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

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

[latex] \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} [/latex] (g)

6 Use the direct shear factor [latex]K_{s}[/latex] and previously assumed values to find the mean stress* [latex] au _{m}[/latex]:

[latex] au_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 [/latex] (h)

7 Find the Wahl factor [latex]K_{w}[/latex] and use it to calculate the alternating shear stress [latex] au _{a}[/latex] in the coil.

[latex] 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 [/latex] (i)

[latex] au_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 [/latex] (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.

[latex] S_{u t} \cong A d^b [/latex] (14.3)

[latex] S_{u s} \cong 0.67 S_{u t} [/latex] (14.4)

[latex] \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} [/latex] (k)

[latex] S_{y s}=0.45 S_{u t}=0.45(244633)=110094 psi [/latex] (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).

[latex] S_{m s} \cong 0.9 S_{u s} \cong 0.9\left(0.67 S_{u t} ight) \cong 0.6 S_{u t} [/latex] (14.14)

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

[latex] S_{e w}=45000 psi [/latex] (m)

[latex] 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 [/latex] (n)

10 The fatigue safety factor for the coils in torsion is calculated from equation 14.18a (p. 814).

[latex] N_{f_s}=\frac{S_{e s}\left(S_{u s}- au_i ight)}{S_{e s}\left( au_m- au_i ight)+S_{u s} au_a} [/latex] (14.18a)

[latex] \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 ight)}=\frac{4(9)^2-(9)-1}{4(9)(9-1)}=1.09 \end{array} [/latex] (o)

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

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):

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

[latex] ext {were} \quad K_b=\frac{4 C_1^2-C_1-1}{4 C_1\left(C_1-1 ight)} [/latex] (14.24b)

[latex] ext {and} \quad C_1=\frac{2 R_1}{d} [/latex] (14.24c)

[latex] \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 ight)}=\frac{4(9)^2-(9)-1}{4(9)(9-1)}=1.09 \end{array} [/latex] (p)

[latex] \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} [/latex] (q)

[latex] \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 [/latex] (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:

[latex] S_{u s} \cong 0.67 S_{u t} [/latex] (14.4)

[latex] N_{f_s}=\frac{S_{e s}\left(S_{u s}- au_i ight)}{S_{e s}\left( au_m- au_i ight)+S_{u s} au_a} [/latex] (14.18a)

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

[latex] \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 } ight)}{S_e\left(\sigma_{ ext {mean }}-\sigma_{\min } ight)+S_{u t} \sigma_{ ext {alt }}} \ &=\frac{38925(244633-81790)}{38925(110416-81790)+244633(28626)}=0.78 \end{aligned} [/latex] (s)

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

[latex] au_B=K_{w_2} \frac{8 D F}{\pi d^3} [/latex] (14.25a)

[latex] ext {where} \quad K_{W_2}=\frac{4 C_2-1}{4 C_2-4} [/latex] (14.25b)

[latex] ext {and} \quad C_2=\frac{2 R_2}{d} [/latex] (14.25c)

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

[latex] \begin{aligned} au_{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 \ au_{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 \ au_{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} [/latex] (u)

14 The hook’s fatigue safety factor in torsion is calculated from equation 14.18a (p. 814).

[latex] N_{f_s}=\frac{S_{e s}\left(S_{u s}- au_i ight)}{S_{e s}\left( au_m- au_i ight)+S_{u s} au_a} [/latex] (14.18a)

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

[latex] \begin{aligned} N_{f_s} &=\frac{S_{e s}\left(S_{u s}- au_{\min } ight)}{S_{e s}\left( au_m- au_{\min } ight)+S_{u s} au_a} \ &=\frac{26080(163918-43435)}{26080(58637-43435)+163918(15202)}=1.1 \end{aligned} [/latex] (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 [latex]C[/latex] 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 [latex]N_{f}[/latex] = 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

[latex] \begin{array}{l} C=7.5 \quad D=1.44 ext { in } \quad K_w=1.20 \quad K_s=1.07\ au_i=15481 psi \quad au_{\min }=27631 psi \quad au_a=10856 psi \quad au_m=37302 psi\ N_{s_{ ext {coil }}}=2.3 \quad N_{f_{s_{c o i l}}}=1.8 \quad N_{f_{s_{ ext {hook }}}}=1.7 \quad N_{f_{b_{ ext {hook }}}}=1.2 \end{array} [/latex] (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.

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

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

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

[latex] k=\frac{d^4 G}{8 D^3 N_a} \quad ext { 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} [/latex] (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 [latex]k[/latex] = 70.7 lb/in.

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

[latex] \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} [/latex] (z)

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

[latex] L_f=L_b+2 L_{ ext {hook }}=1.97+2(1.25)=4.46 ext { in } [/latex] (aa)

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

[latex] y_{\max }=\frac{F_{ ext {max }}-F_{ ext {initial }}}{k}=\frac{85-28}{70.7}=0.81 ext { in } [/latex] (ab)

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

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

[latex] 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 [/latex] (ac)

The ratio between the natural frequency and the forcing frequency is

[latex] \frac{8440}{500}=16.9 [/latex] (ad)

which is sufficiently high.

23 The design specification for this A228 wire spring is

[latex] d=0.192 ext { in } \quad O D=1.63 ext { in } \quad N_t=10.25 \quad L_f=4.46 [/latex] (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 [latex]K_{s}[/latex] 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.

Table 14-14a Example 14-5—Design of a Helical Extension Spring for Cyclic Loads Part 1 of 2
Input	Variable	Output	Unit	Comments
500	[latex] fn [/latex]		rpm	excitation frequency
7.50	[latex] C [/latex]			trial spring index
0.192	[latex] d [/latex]		in	trial wire diameter (List Function)
0.50	[latex] y [/latex]		in	deflection bet Fmin & Fmax
	[latex] ymax [/latex]	0.81	in	maximum deflection at Fmax
	[latex] ymin [/latex]	0.31	in	minimum deflection at Fmin
'music	[latex] matl [/latex]			'music, 'oiltemp, 'hdrawn, etc.
85	[latex] Fmax [/latex]		lb	maximum applied force
50	[latex] Fmin [/latex]		lb	minimum applied force
	[latex] Finit [/latex]	28.01	lb	force from initial tension
	[latex] Flow [/latex]	50.00	lb	lowest applied force on spring
	[latex] Falt [/latex]	17.50	lb	alternating force
	[latex] Fmean [/latex]	67.50	lb	mean force
	[latex] k [/latex]	70.70	lb/in	spring rate with [latex]N_{a}[/latex] rounded
	[latex] N [/latex]	9.35		no. of active coils—exact
	[latex] Na [/latex]	9.25		no. of active coils—nearest 1/4 coil
	[latex] Ntot [/latex]	10.25		no. of total coils
	[latex] D [/latex]	1.44	in	mean coil diameter
	[latex] Dout [/latex]	1.63	in	outside coil diameter
	[latex] Din [/latex]	1.25	in	inside coil diameter
	[latex] Ks [/latex]	1.07		static factor—Eq. 13.8
	[latex] Kw [/latex]	1.20		Wahl factor—Eq. 13.9
	[latex] tauinit [/latex]	15481	psi	shear stress from initial tension
	[latex] taumin [/latex]	27631	psi	shear stress at Fmin
	[latex] taumax [/latex]	46973	psi	shear stress at Fmax
	[latex] taualt [/latex]	10856	psi	alternating shear stress for fatigue
	[latex] taumean [/latex]	37302	psi	mean shear stress for fatigue
	[latex] Sut [/latex]	241441	psi	tensile strength—Eq. 13.3 & Table 13-4
	[latex] Sus [/latex]	161765	psi	ultimate shear strength—Eq. 13.4
	[latex] Ssy [/latex]	108648	psi	shear yield based on Table 13-10
	[latex] Ssyh [/latex]	96576	psi	shear yield in hook—Table 13-10
	[latex] Sew [/latex]	45000	psi	wire endurance limit—Eq. 13.12
	[latex] Ses [/latex]	26135	psi	fully reversed endr. limit—Eq. 13.16b
	[latex] Sy [/latex]	181081	psi	yield strength in bending
	[latex] Se [/latex]	39008	psi	endurance limit in tension

Table 14-14b Example 14-5—Design of a Helical Extension Spring for Cyclic Loads Part 2 of 2
Input	Variable	Output	Unit	Comments
	[latex] N_{f} [/latex]	1.8		SF coils—fatigue—Eq. 13.14
	[latex] N_{s} [/latex]	2.3		SF coils—static loading at Fmax
	[latex] N_{fht} [/latex]	1.7		SF hook—torsion fatigue
	[latex] N_{sht} [/latex]	1.8		SF hook—torsional yielding
	[latex] N_{fhs} [/latex]	1.2		SF hook—bending fatigue
	[latex] N_{shs} [/latex]	1.8		SF hook—bending yielding
	[latex] L_{body} [/latex]	1.97	in	length of coil body
	[latex] hook1 [/latex]	1.25	in	length of hook on one end
	[latex] hook2 [/latex]	1.25	in	length of hook on other end
	[latex] L_{f} [/latex]	4.46	in	free length inside hooks
	[latex] W_{total} [/latex]	0.38	lb	weight of total coils—Eq. 13-11b
	[latex] n_{f} [/latex]	140.7	Hz	natural frequency in Hz
	[latex] c_{rpm} [/latex]	8440	rpm	natural frequency in rpm
	[latex] FreqFac [/latex]	16.9		ratio—nat. freq. to forcing freq.
500	[latex] C_{2} [/latex]			should be set > 4
	[latex] R_{2} [/latex]	0.48	in	side bend radius at hook root
	[latex] K_{hook} [/latex]	1.19		[latex]K[/latex] factor for hook in torsion
	[latex] t_{maxhook} [/latex]	52294	psi	maximum torsional stress in hook
	[latex] t_{minhook} [/latex]	30761	psi	minimum torsional stress in hook
	[latex] t_{althook} [/latex]	10766	psi	alternating torsional stress in hook
	[latex] t_{mnhook} [/latex]	41528	psi	mean torsional stress in hook
	[latex] C_{1} [/latex]	7.50		spring index for hook

Table 14-4 Coefficients and Exponents for Equation 14.3 Source: Reference 1
ASTM#	Material	Range		Exponent [latex]b[/latex]	Coefficient [latex]A[/latex]		Correlation Factor
ASTM#	Material	mm	in	Exponent [latex]b[/latex]	MPa	psi	Correlation Factor
A227	Cold drawn	0.5–16	0.020–0.625	–0.182 2	1 753.3	141 040	0.998
A228	Music wire	0.3–6	0.010–0.250	–0.1625	2 153.5	184 649	0.9997
A229	Oil tempered	0.5–16	0.020–0.625	–0.183 3	1 831.2	146 780	0.999
A232	Chrome-v	0.5–12	0.020–0.500	–0.145 3	1 909.9	173 128	0.998
A401	Chrome-s.	0.8–11	0.031–0.437	–0.093 4	2 059.2	220 779	0.991

Table 14-12 Maximum Torsional and Bending Yield Strengths [latex]S_{ys}[/latex] and [latex]S_{y}[/latex] for Helical Extension Springs in Static Applications No Set Removal and Low-Temperature Heat Treatment Applied. Source: Ref. 1
Material	Maximum Percent of Ultimate Tensile Strength
	[latrex]S_{ys}[/latex] in Torsion		[latex]S_{y}[/latex] in Bending
	Body	End	End
Cold-drawn carbon steel (e.g., A227, A228)	45%	40%	75%
Hardened and tempered carbon and low-alloy steel (e.g., A229, A230, A232, A401)	50	40	75
Austenitic stainless steel and nonferrous alloys (e.g., A313, B134, B159, B197)	35	30	55

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