Question 19.EP.4: A helical extension spring is to be installed in a latch for...

As before, the suggested design procedure will be given as numbered steps, followed by the calculations for this set of data.

Step 1. Assume a trial mean diameter and a trial design stress for the spring. Let the mean diameter be 0.500 in. For ASTM A227 wire under average service, a design stress of 110 000 psi is reasonable (from Figure 19_8).

Step 2. Compute a trial wire diameter from Equation (19_4)  \tau=\frac{8 K F C}{\pi D_{w}^{2}} for the maximum operating force, the assumed mean diameter and design stress, and an assumed value for K of about 1.20. D_{w}=\left[\frac{8 K F_{o} D_{m}}{\pi \tau_{d}}\right]^{1 / 3}=\left[\frac{(8)(1.20)(26.75)(0.50)}{(\pi)(110000)}\right]^{1 / 3}=0.072 \text { in } A standard wire size in the U.S. Steel Wire Gage system is available in this size. Use 15-gage.

Step 3. Determine the actual design stress for the selected wire size. From Figure 19_8. at a wire size of 0.072 in, the design stress is 120 000 psi.

Step 4. Compute the actual values for outside diameter, mean diameter, inside diameter, spring index, and the Wahl factor, K. These factors are the same as those defined for helical compression  springs. Let the outside diameter be as specified, 0.625 in. Then D_{m}=O D-D_{11}=0.625-0.072=0.553 \text { in }

Step 5. Compute the actual expected stress in the spring wire under the operating load from Equation (19_4):\tau=\frac{8 K F C}{\pi D_{w}^{2}}

\tau_{o}=\frac{8 K F_{o} D_{m}}{\pi D_{w}^{3}}=\frac{8(1.19)(26.75)(0.553)}{\pi(0.072)^{3}}=120000 \mathrm{psi} (okay)

Step 6. Compute the required number of coils to produce the desired deflection characteristics. Solve Equation (19_6)  f=\frac{8 F D_{m}^{3} N_{a}}{G D_{w}^{4}}=\frac{8 F C^{3} N_{a}}{G D_{w}} for the number of coils, and substitute k = force/deflection = F/f. k=\frac{26.75-16.25}{4.25-3.50}=14.0 \mathrm{lb} / \text { in }

Step 7. Compute the body length for the spring, and propose a trial design for the ends. \text { Body length }=D_{w}\left(N_{a}+1\right)=(0.072)(16.3+1)=1.25 \mathrm{in} Let’s propose to use a full loop at each end of the spring, adding a length equal to the ID of
the .spring at each end. Then the total free length is L_{f}=\text { body length }+2(I D)=1.25+2(0.481)=2.21 \text { in }

Step 8. Compute the deflection from free length to operating length: f_{o}=L_{o}-L_{f}=4.25-2.21=2.04 \mathrm{in}

Step 9. Compute the initial force in the spring at which the coils just begin to separate. This is done by subtracting the amount of force due to the deflection,f_{o}: F_{1}=F_{o}-k f_{o}=26.75-(14.0)(2.04)=-1.81 \mathrm{lb} The negative force resulting from a free length that is too small for the specified conditions is clearly impossible. Let’s tryL_{f} = 2.50 in, which will require a redesign of the end loops. Then f_{\sigma}=4.25-2.50=1.75 \mathrm{in}

F_{l}=26.75-(14.0)(1.75)=2.25 \mathrm{lb} (reasonable)

Step 10. Compute the stress in the spring under the initial tension, and compare it with the recommended levels in Figure 19_21. Because the stress is proportional to the load, \tau_{l}=\tau_{o}\left(F_{l} / F_{o}\right)=(120000)(2.25 / 26.75)=10100 \mathrm{psi} For C = 7.68, this .stress is in the preferred range from Figure 19_21. At this point, the coiled portion of the spring is satisfactory. The final configuration of the end loops must be completed and analyzed for stress.