Design of a Helical Torsion Spring for Cyclic Loading Problem Design a torsion spring for a dynamic load over a given deflection. Given The spring must give a minimum moment of 50 lb-in and a maximum moment of 80 lb-in over a dynamic deflection of 0.25 revolutions (90°). An infinite life is desired

Question

Design of a Helical Torsion Spring for Cyclic Loading

Problem Design a torsion spring for a dynamic load over a given deflection.

Given The spring must give a minimum moment of 50 lb-in and a maximum moment of 80 lb-in over a dynamic deflection of 0.25 revolutions (90°). An infinite life is desired.

Assumptions Use unpeened music wire (ASTM A228). Use 2-in long straight tangent ends. The coil is loaded to close it.

Accepted Answer

See Figure 14-27.

1 Assume a 0.192-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(0.192)=1.73 ext { in } [/latex] (a)

2 Find the mean and alternating moments:

[latex] \begin{array}{c} M_m=\frac{M_{\max }+M_{\min }}{2}=\frac{80+50}{2}=65 lb \ M_a=\frac{M_{\max }-M_{\min }}{2}=\frac{80-50}{2}=15 lb \end{array} [/latex] (b)

3 Find the Wahl bending factor for the inside surface [latex]K_{b_i}[/latex] and use it to calculate the maximum compressive stress in the coil at the inner surface.

[latex] K_{b_i}=\frac{4 C^2-C-1}{4 C(C-1)}=\frac{4(9)^2-9-1}{4(9)(9-1)}=1.090 [/latex] (c)

[latex] \sigma_{i_{\max }}=K_{b_i} \frac{32 M_{\max }}{\pi d^3}=1.09 \frac{32(80)}{\pi(0.192)^3}=125523 psi [/latex] (d)

4 Find the Wahl bending factor [latex]K_{b_o}[/latex] for the outside surface and calculate the maximum, minimum, alternating, and mean tensile stresses in the coil at the outer surface.

[latex] K_{b_o}=\frac{4 C^2+C-1}{4 C(C+1)}=\frac{4(9)^2+9-1}{4(9)(9+1)}=0.9222 [/latex] (e)

[latex] \begin{array}{l} \sigma_{o_{\min }}=K_{b_o} \frac{32 M_{\min }}{\pi d^3}=0.9222 \frac{32(50)}{\pi(0.192)^3}=66359 psi \ \sigma_{o_{\max }}=K_{b_o} \frac{32 M_{\max }}{\pi d^3}=0.9222 \frac{32(80)}{\pi(0.192)^3}=106175 psi \end{array} [/latex] (f)

[latex] \begin{aligned} \sigma_{o_{ ext {mean }}} &=\frac{\sigma_{o_{\max }}+\sigma_{o_{\min }}}{2}=\frac{66359+106175}{2}=86267 \ \sigma_{o_{ ext {alt }}} &=\frac{\sigma_{o_{\max }}-\sigma_{o_{\min }}}{2}=\frac{66359-106175}{2}=19908 \end{aligned} [/latex] (g)

5 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 bending yield strength for the coil body from Table 14-15, assuming no stress relieving.

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

[latex] S_{u t}=A d^b=184649(0.192)^{-0.1625}=241441 psi [/latex] (h)

[latex] S_y=1.0 S_{u t}=241441 psi [/latex] (i)

6 Find the wire bending endurance limit for unpeened springs from equation 14.34 and convert it to a fully reversed bending endurance strength with equation 14.35b.

[latex] S_{e w_b^{\prime}}=\frac{S_{e w^{\prime}}}{0.577} [/latex] (14.34a)

[latex] \begin{array}{l} S_{e w_b^{\prime}} \cong \frac{45.0}{0.577}=78 kpsi \quad(537 MPa ) \quad ext {for unpeened springs}\ S_{e w_b^{\prime}} \cong \frac{67.5}{0.577}=117 kpsi (806 MPa ) \quad ext {for peened springs} \end{array} [/latex] (14.43b)

[latex] N_{f_b}=\frac{S_e\left(S_{u t}-\sigma_{o_{ ext {min }}} ight)}{S_e\left(\sigma_{o_{ ext {mean }}}-\sigma_{o_{ ext {min }}} ight)+S_{u t} \sigma_{o_{ ext {alt }}}} [/latex] (14.35b)

[latex] S_{e w_b^{\prime}} \cong \frac{45000}{0.577}=77990 psi [/latex] (j)

[latex] S_e=0.5 \frac{S_{e w_b} S_{u t}}{S_{u t}-0.5 S_{e w_b}}=0.5 \frac{77990(241441)}{241441-0.5(77990)}=46506 psi [/latex] (k)

7 The fatigue safety factor for the coils in bending is calculated from equation 14.35a.

[latex] N_y=\frac{S_y}{\sigma_{i_{\max }}} [/latex] (14.35a)

[latex] \begin{aligned} N_{f_b} &=\frac{S_e\left(S_{u t}-\sigma_{o_{ ext {min }}} ight)}{S_e\left(\sigma_{o_{ ext {mean }}}-\sigma_{o_{ ext {min }}} ight)+S_{u t} \sigma_{o_{ ext {alt }}}} \ &=\frac{46506(241441-66359)}{46506(86267-66359)+241441(19908)}=1.4 \end{aligned} [/latex] (l)

8 The static safety factor against yielding is

[latex] N_{y_b}=\frac{S_y}{\sigma_{i_{\max }}}=\frac{241441}{125523}=1.9 [/latex] (m)

These are both acceptable safety factors.

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

[latex] k=\frac{\Delta M}{ heta}=\frac{80-50}{0.25}=120 lb - in / rev [/latex] (n)

10 To get the defined spring rate, the number of active coils must satisfy equation 14.29:

[latex] k=\frac{M}{ heta_{ ext {rev }}} \cong \frac{d^4 E}{10.8 D N_a} [/latex] (14.29)

[latex] k=\frac{d^4 E}{10.8 D N_a} \quad ext { or } \quad N_a=\frac{d^4 E}{10.8 D k}=\frac{(0.192)^4 30 E 6}{10.8(1.73)(120)}=18.2 [/latex] (o)

The ends contribute to the active coils as

[latex] N_e=\frac{L_1+L_2}{3 \pi D}=\frac{2+2}{3 \pi(1.73)}=0.25 [/latex] (p)

and the number of body coils in the spring is

[latex] N_b=N_a-N_e=18.2-0.25 \cong 18 [/latex] (q)

11 The angular deflections at the specified loads from equation 14.28c are

[latex] heta_{r e v} \cong 10.8 \frac{M D N_a}{d^4 E} [/latex] (14.28c)

[latex] heta_{\min } \cong 10.8 \frac{M_{\min } D N_a}{d^4 E}=10.8 \frac{50(1.73)(18.2)}{(0.192)^4(30 E 6)}=0.417 rev =150 deg [/latex] (r)

[latex] heta_{\max } \cong 10.8 \frac{M_{\max } D N_a}{d^4 E}=10.8 \frac{80(1.73)(18.2)}{(0.192)^4(30 E 6)}=0.667 rev =240 deg [/latex] (s)

12 The files EX14-06 can be found on the CD-ROM.

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 14-15 Maximum Recommended Bending Yield Strength [latex]S_{y}[/latex] for Helical Torsion Springs in Static Applications Source: Adapted from Reference 1
Materia	Maximum Percent of Ultimate Tensile Strength
Materia	Stress Relieved	Favorable Residual Stress
Cold-drawn carbon steel (e.g., A227, A228)	80%	100%
Hardened and tempered carbon and low-alloy steel (e.g., A229, A230, A232, A401)	85	100
Austenitic stainless steel and nonferrous alloys (e.g., A313, B134, B159, B197)	60	80

Question 14.6: Design of a Helical Torsion Spring for Cyclic Loading Proble...

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