The solution is presented in the same order as the requested items just listed. The formulas used are found in the preceding sections of this chapter.
Step 1. The wire is 24-gage music wire (Table 19_2).
Gage
no. |
U.S. Steel
Wire Gage
\left(in\right) ^{a} |
Music Wire
Gage \left(in\right) ^{b} |
Brown &
Sharpe Gage
\left(in\right) ^{c} |
Preferred
metric diameters
\left(mm\right) ^{d} |
| 23 |
0.0258 |
0.051 |
0.0226 |
0.65 |
| 24 |
0.0230 |
0.055 |
0.0201 |
0.60 or 0.55 |
| 25 |
0.0204 |
0.059 |
0.0179 |
0.50 or 0.55 |
| 26 |
0.0181 |
0.063 |
0.0159 |
0.45 |
| 27 |
0.0173 |
0.067 |
0.0142 |
0.45 |
| 28 |
0.0162 |
0.071 |
0.0126 |
0.40 |
| 29 |
0.0150 |
0.075 |
0.0113 |
0.40 |
| 30 |
0.0140 |
0.080 |
0.0100 |
0.35 |
| 31 |
0.0132 |
0.085 |
0.00893 |
0.35 |
| 32 |
0.0128 |
0.090 |
0.00795 |
0.30 or 0.35 |
| 33 |
0.0118 |
0.095 |
0.00708 |
0.30 |
| 34 |
0.0104 |
0.100 |
0.00630 |
0.28 |
| 35 |
0.0095 |
0.106 |
0.00501 |
0.25 |
| 36 |
0.0090 |
0.112 |
0.00500 |
0.22 |
| 37 |
0.0085 |
0.118 |
0.00445 |
0.22 |
| 38 |
0.0080 |
0.124 |
0.00396 |
0.20 |
| 39 |
0.0075 |
0.130 |
0.00353 |
0.20 |
| 40 |
0.007 |
0.138 |
0.00314 |
0.18 |
Thus, D_{m}=O D-D_{w}=0.561-0.055=0.506 \text { in }
I D=D_{m}-D_{w}=0.506-0.055=0.451 \text { in }
\text { Spring index }=C=D_{m} / D_{w}=0.506 / 0.055=9.20
\text { Wahl factor }=K=(4 C-1) /(4 C-4)+0.615 / C
K = [4(9.20) – l]/[4(9.20) – 4] + 0.615/9.20
K = 1.158
Step 2. Stress in spring at F = F_{a} = 14.0 Ib [Equation (19_4)]:
\tau=\frac{8 K F D_{m}}{\pi D_{w}^{3}}=\frac{8 K F C}{\pi D_{w}^{2}}
\tau_{o}=\frac{8 K F_{o} C}{\pi D_{\mathrm{w}}^{2}}=\frac{8(1.158)(14.0)(9.20)}{\pi(0.055)^{2}}=125560 \mathrm{psi}
Step 3. Deflection at operating force [Equation (19_6)]: f=\frac{8 F D_{m}^{3} N_{a}}{G D_{\mathrm{w}}^{4}}=\frac{8 F C^{3} N_{a}}{G D_{w}}
f_{o}=\frac{8 F_{o} C^{3} N_{a}}{G D_{w}}=\frac{8(14.0)(9.20)^{3}(8.0)}{\left(11.85 \times 10^{6}\right)(0.055)}=1.071 \mathrm{in}
Note that the number of active coils for a spring with squared and ground ends is N_{a}=N-2=10.0 – 2 = 8.0. Also, the spring wire modulus, G, was found in Table 19_4.
|
Shear modulus,G |
|
Tension modulus,E |
|
Material
and ASTM no. |
(psi) |
(GPa) |
(psi) |
(GPa) |
| Hard-drawn steel: A227 |
11.5 ×10^{6} |
79.3 |
28.6 ×10^{6} |
197 |
| Music wire: A228 |
11.85 ×10^{6} |
81.7 |
29.0 ×10^{6} |
200 |
| Oil-tempered: A229 |
11.2 ×10^{6} |
77.2 |
28.5 ×10^{6} |
196 |
| Chromium-vanadium: A231 |
11.2 ×10^{6} |
77.2 |
28.5×10^{6} |
196 |
| Chromium-silicon: A40I |
11.2 ×10^{6} |
77.2 |
29.5× 10^{6} |
203 |
| Stainless steels: A313 |
|
|
|
|
| Types302, 304, 316 |
10.0 ×10^{6} |
69.0 |
28.0 × 10^{6} |
193 |
| Type 17-7 PH |
10.5 ×10^{6} |
72.4 |
29.5 × 10^{6} |
203 |
| Spring brass: B134 |
5.0 ×10^{6} |
34.5 |
15.0 × 10^{6} |
103 |
| Phosphor bronze: B159 |
6.0 ×10^{6} |
41.4 |
15.0× 10^{6} |
103 |
| Beryllium copper: B197 |
7.0 ×10^{6} |
48,3 |
17.0 × 10^{6} |
117 |
| Monel and K-Monel |
9.5 ×10^{6} |
65.5 |
26.0 × 10^{6} |
179 |
| Inconel and Inconel-X |
10.5 ×10^{6} |
72.4 |
31.0 × 10^{6} |
214 |
The value of f_{a} is the deflection from free length to the operating length.
Step 4. Operating length: We compute operating length as
L_{o}=L_{f}-f_{o}=1.75-1.071=0.679 \mathrm{in}
\text { Solid length }=L_{s}=D_{w}(N)=0.055(10.0)=0.550 \mathrm{in} Spring Index: We use Equation (19_1). k=\Delta F / \Delta L
k=\frac{\Delta F}{\Delta L}=\frac{F_{o}}{L_{f}-L_{o}}=\frac{F_{o}}{f_{o}}=\frac{14.0 \mathrm{lb}}{1.071 \mathrm{in}}=13.07 \mathrm{lb} / \mathrm{in}
Step 5. We can find the force at solid length by multiplying the spring rate times the deflection from the free length to the solid length. Then
F_{s}=k\left(L_{f}-L_{s}\right)=(13.07 \mathrm{lb} / \mathrm{in})(1.75 \mathrm{in}-0.550 \mathrm{in})=15.69 \mathrm{lb}
The stress at solid length, \tau _{s}, could be found from Equation (19_4), \tau=\frac{8 K F D_{m}}{\pi D_{w}^{3}}=\frac{8 K F C}{\pi D_{w}^{2}} using F = F_{s}. However, an easier method is to recognize that the stress is directly proportional to the force on the spring and that all of the other data in the formula are the same as those used to compute the stress under the operating force, F_{o}. We can then use the simple proportion
\tau_{s}=\tau_{o}\left(F_{s} / F_{o}\right)=(125560 \mathrm{psi})(15.69 / 14.0)=140700 \mathrm{psi}
Step 6. Design stress,\tau _{d}: From Figure 19_9, in the graph of design stress versus spring wire diameter for ASTM A228 steel, we can use the average service curve based on the expected number of cycles of loading. We read \tau _{d} = 135 000 psi for the 0.055-in wire. Because the actual operating stress,\tau _{o}, is less than this value, it is satisfactory.
Step 7. Maximum allowable stress,\tau _{\max } : It is recommended that the light service curve be used to determine this value. For D _{W } = 0.055, \tau _{\max } = 150 000 psi. The actual expected maximum stress that occurs at solid length (\tau _{s } = 140 700 psi) is less than this value, and therefore the design is satisfactory with regard to stresses.
Step 8. Buckling: To evaluate buckling, we must compute
L_{f} / D_{m}=(1.75 \mathrm{in}) /(0.506 \mathrm{in})=3.46
Referring to Figure 19_15 and using curve A for squared and ground ends, we see that the critical deflection ratio is very high and that buckling should not occur. In fact, for any value of L_{f} / D_{m}<5.2,we can conclude that buckling will not occur.
Coil clearance, cc: We evaluate cc as follows:
c c=\left(L_{o}-L_{s}\right) / N_{a}=(0.679-0.550) /(8.0)=0.016 \text { in }
Comparing this to the recommended minimum clearance of
D_{w} / 10=(0.055 \mathrm{in}) / 10=0.0055 \mathrm{in}
we can judge this clearance to be acceptable.
Step 9. Hole diameter: It is recommended that the hole into which the spring is to be installed should be greater in diameter than the OD of the spring by the amount of D_{w} / 10. Then D_{\text {hole }}>O D+D_{\mathrm{w}} / 10=0.561 \text { in }+(0.055 \mathrm{in}) / 10=0.567 \mathrm{in}
A diameter of 5/8 in (0.625 in) would be a satisfactory standard size. This completes the example problem.
