Question 14.6: Design of a Helical Torsion Spring for Cyclic Loading Proble...
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.
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 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(0.192)=1.73 \text { in } (a)
2 Find the mean and alternating moments:
\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} (b)
3 Find the Wahl bending factor for the inside surface K_{b_i} and use it to calculate the maximum compressive stress in the coil at the inner surface.
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 (c)
\sigma_{i_{\max }}=K_{b_i} \frac{32 M_{\max }}{\pi d^3}=1.09 \frac{32(80)}{\pi(0.192)^3}=125523 psi (d)
4 Find the Wahl bending factor K_{b_o} for the outside surface and calculate the maximum, minimum, alternating, and mean tensile stresses in the coil at the outer surface.
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 (e)
\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} (f)
\begin{aligned} \sigma_{o_{\text {mean }}} &=\frac{\sigma_{o_{\max }}+\sigma_{o_{\min }}}{2}=\frac{66359+106175}{2}=86267 \\ \sigma_{o_{\text {alt }}} &=\frac{\sigma_{o_{\max }}-\sigma_{o_{\min }}}{2}=\frac{66359-106175}{2}=19908 \end{aligned} (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.
S_{u t} \cong A d^b (14.3)
S_{u t}=A d^b=184649(0.192)^{-0.1625}=241441 psi (h)
S_y=1.0 S_{u t}=241441 psi (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.
S_{e w_b^{\prime}}=\frac{S_{e w^{\prime}}}{0.577} (14.34a)
\begin{array}{l} S_{e w_b^{\prime}} \cong \frac{45.0}{0.577}=78 kpsi \quad(537 MPa ) \quad \text {for unpeened springs}\\ S_{e w_b^{\prime}} \cong \frac{67.5}{0.577}=117 kpsi (806 MPa ) \quad \text {for peened springs} \end{array} (14.43b)
N_{f_b}=\frac{S_e\left(S_{u t}-\sigma_{o_{\text {min }}}\right)}{S_e\left(\sigma_{o_{\text {mean }}}-\sigma_{o_{\text {min }}}\right)+S_{u t} \sigma_{o_{\text {alt }}}} (14.35b)
S_{e w_b^{\prime}} \cong \frac{45000}{0.577}=77990 psi (j)
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 (k)
7 The fatigue safety factor for the coils in bending is calculated from equation 14.35a.
N_y=\frac{S_y}{\sigma_{i_{\max }}} (14.35a)
\begin{aligned} N_{f_b} &=\frac{S_e\left(S_{u t}-\sigma_{o_{\text {min }}}\right)}{S_e\left(\sigma_{o_{\text {mean }}}-\sigma_{o_{\text {min }}}\right)+S_{u t} \sigma_{o_{\text {alt }}}} \\ &=\frac{46506(241441-66359)}{46506(86267-66359)+241441(19908)}=1.4 \end{aligned} (l)
8 The static safety factor against yielding is
N_{y_b}=\frac{S_y}{\sigma_{i_{\max }}}=\frac{241441}{125523}=1.9 (m)
These are both acceptable safety factors.
9 The spring rate is defined from the two specified moments at their relative deflection.
k=\frac{\Delta M}{\theta}=\frac{80-50}{0.25}=120 lb – in / rev (n)
10 To get the defined spring rate, the number of active coils must satisfy equation 14.29:
k=\frac{M}{\theta_{\text {rev }}} \cong \frac{d^4 E}{10.8 D N_a} (14.29)
k=\frac{d^4 E}{10.8 D N_a} \quad \text { 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 (o)
The ends contribute to the active coils as
N_e=\frac{L_1+L_2}{3 \pi D}=\frac{2+2}{3 \pi(1.73)}=0.25 (p)
and the number of body coils in the spring is
N_b=N_a-N_e=18.2-0.25 \cong 18 (q)
11 The angular deflections at the specified loads from equation 14.28c are
\theta_{r e v} \cong 10.8 \frac{M D N_a}{d^4 E} (14.28c)
\theta_{\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 (r)
\theta_{\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 (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 b |
Coefficient A | Correlation Factor | ||
| mm | in | MPa | psi | ||||
| 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 S_{y} for Helical Torsion Springs in Static Applications Source: Adapted from Reference 1 |
||
| Materia | Maximum Percent of Ultimate Tensile Strength | |
| 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 |
