A timer incorporates a mechanism to close a switch after the timer rotates one complete revolution. The switch contacts are actuated by a torsion spring that is to be designed. A cam on the timer shaft slowly moves a lever attached to one end of the spring to a point where the maximum torque on the spring is 3.00 Ib·in. At the end of the revolution, the cam permits the lever to rotate 60° suddenly with the motion produced by the energy stored in the spring. At this new position, the torque on the spring is 1.60 Ib·in. Because of space limitations, the OD of the spring should not be greater than 0.50 in,and the length should not be greater than 0.75 in. Use music wire, ASTM A228 steel wire. The number of cycles of the spring will be moderate, so use design stresses for average service.
Question 19.EP.5: A timer incorporates a mechanism to close a switch after the...
Step 1. Assume a trial value for the mean diameter and an estimate for the design stress. Let’s use a mean diameter of 0.400 in and estimate the design stress for A228 music wire, average service, to be 180 000 psi (Figure 19_25).
Step 2. Solve Equation (19_19) \sigma=\frac{M c K_{b}}{I}=\frac{M K_{b}}{S}=\frac{M K_{b}}{\pi D_{w}^{3} / 32}=\frac{32 M K_{b}}{\pi D_{w}^{3}} for the wire diameter, compute a trial size, and select a standard wire size. Let K_{b}=1.15 as an estimate. Also use the largest applied torque: D_{w}=\left[\frac{32 M K_{b}}{\pi \sigma_{d}}\right]^{1 / 3}=\left[\frac{32(3.0)(1.15)}{\pi(180000)}\right]^{1 / 3}=0.058 \text { in } From 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 |
we can choose 25-gage music wire with a diameter of 0.059 in. For this size wire, the actual design stress for average service is 178 000 psi.
Step 3. Compute the OD, ID, spring index, and the new K_{b} : O D=D_{m}+D_{w}=0.400+0.059=0.459 \text { in (okay) }
I D=D_{m}-D_{w}=0.400-0.059=0.341 \text { in }
C=D_{m} / D_{w}=0.400 / 0.059=6.78
K_{b}=\frac{4 C^{2}-C-1}{4 C(C-1)}=\frac{4(6.78)^{2}-6.78-1}{4(6.78)(6.78-1)}=1.123
Step 4. Compute the actual expected stress from Equation (19_19): \sigma=\frac{M c K_{b}}{I}=\frac{M K_{b}}{S}=\frac{M K_{b}}{\pi D_{w}^{3} / 32}=\frac{32 M K_{b}}{\pi D_{w}^{3}}
\sigma=\frac{32 M K_{b}}{\pi D_{w}^{3}}=\frac{32(3.0)(1.123)}{(\pi)(0.059)^{3}}=167000 \mathrm{psi} \quad \text { (okay) }
Step 5. Compute the spring rate from the given data. The torque exerted by the spring decreases from 3.00 to 1.60 Ib·in as the spring rotates 60°. Convert 60° to a fraction of a revolution (rev):\theta=\frac{60}{360}=0.167 \mathrm{rev}
k_{\theta}=\frac{M}{\theta}=\frac{3.00-1.60}{0.167}=8.40 \mathrm{lb} \cdot \mathrm{in} / \mathrm{rev}
Step 6. Compute the required number of coils by solving forN_{a} from Equation (19_22):k_{\theta}=\frac{M}{\theta}=\frac{E D_{w}^{4}}{10.2 D_{m} N_{a}}
N_{a}=\frac{E D_{w}^{4}}{10.2 D_{m} k_{\theta}}=\frac{\left(29 \times 10^{6}\right)(0.059)^{4}}{(10.2)(0.400)(8.40)}=10.3 \text { coils }
Step 7. Compute the equivalent number of coils due to the ends of the spring from Equation (19_23).N_{e}=\left(L_{1}+L_{2}\right) /\left(3 \pi D_{m}\right) This requires some design decisions. Let’s use straight ends, 2.0 in long on one side and 1.0 in long on the other. These ends will be attached to the structure of the timer during operation. Then N_{e}=\left(L_{1}+L_{2}\right) /\left(3 \pi D_{m}\right)=(2.0+1.0) /[3 \pi(0.400)]=0.80 \text { coil }
Step 8. Compute the required number of coils in the body of the spring: N_{b}=N_{a}-N_{e}=10.3-0.8=9.5 \text { coils }
Step 9. Complete the geometric design of the spring, including the size of the rod on which it will be mounted.We first need the total angular deflection for the spring from the free condition to the maximum load. In this case, we know that the spring rotates 60° during operation. To this we must add the rotation from the free condition to the initial torque, 1.60 Ib·in. Thus, \theta_{I}=M_{I} / k_{\theta}=1.60 \mathrm{lb} \cdot \mathrm{in} /(8.4 \mathrm{lb} \cdot \mathrm{in} / \mathrm{rev})=0.19 \mathrm{rev} Then the total rotation is \theta_{t}=\theta_{l}+\theta_{o}=0.19+0.167=0.357 \mathrm{rev} From Equation (19_15), \tau_{B}=\frac{8 D_{m} F_{o} K_{2}}{\pi D_{w}^{3}} the mean diameter at the maximum operating torque is D_{m}=D_{m t} N_{a} /\left(N_{a}+\theta_{t}\right)=[(0.400)(10.3)] /[(10.3+0.357)]=0.387 \text { in } The minimum inside diameter is I D_{\min }=0.387-D_{w}=0.387-0.059=0.328 \text { in } The rod diameter on which the spring is mounted should be approximately 0.90 times this value. Then D_{r}=0.9(0.328)=0.295 \text { in } \quad \text { (say. } 0.30 \mathrm{in} \text { ) } The length of the spring, assuming that all coils are originally touching, is computed from Equation (19_18): L=D_{w}\left(N_{a}+1+\theta\right)
L_{\max }=D_{w}\left(N_{a}+1+\theta_{t}\right)=(0.059)(10.3+1+0.356)=0.688 \text { in } \quad \text { (okay) } This value for length is the maximum space required in the direction along the axis of the coil when the spring is fully actuated. The specifications allow an axial length of 0.75 in, so this design is acceptable.
