Question 19.EP.2: A helical compression spring is to exert a force of 8.0 lb w...

Solution Method 1                 The procedure works directly toward the overall geometry of the spring by specifying the mean diameter to meet the space limitations. The process requires that the designer have tables of data available for wire diameters (such as Table 19_2)

and graphs of design stresses for the material from which the spring will be made (such as Figures 19_8 through 19_13). We must make an initial estimate for the design stress for the material by consulting the  charts of design stress versus wire diameter to make a reasonable choice. In general, more than one trial must be made, but the results of early trials will help you decide the values to use for later trials.

Step 1. Specify a material and its shear modulus of elasticity, G. For this problem, several standard spring materials can be used. Let’s select ASTM A231 chromium-vanadium steel wire, having a value of G = 11 200 000 psi (see Table 19_4).

Step 2. From the problem statement, identify the operating force, F_{o} : the operating length at which that force must be exerted,L_{o} ; the force at some other length, called the installed force, F_{i} ; and the installed length L_{i} . Remember, F_{o} is the maximum force that the spring experiences under normal operating conditions. Many times, the second force level is not specified. In that ca.se, let F_{i} = 0, and specify a design value for the free length, L_{f} . in place of L_{i} . For this problem, F_{o}=12.0 \mathrm{lb} ; L_{o}=1.25 \mathrm{in} ; F_{i}=8.0 \mathrm{lb} ; \text { and } L_{i}=1.75 \mathrm{in} .

Step 3. Compute the spring rate, k, using Equation (19_1a)k=\frac{F_{o}-F_{i}}{L_{i}-L_{o}}

Step 4. Compute the free length, L_{f} : L_{f}=L_{i}+F_{i} / k=1.75 \mathrm{in}+[(8.00 \mathrm{lb}) /(8.00 \mathrm{lb} / \mathrm{in})]=2.75 \mathrm{in} The second term in the preceding equation is the amount of deflection from free length to the installed length in order to develop the installed force, f_{i} . Of course, this step becomes unnecessary if the free length is specified in the original data.

Step 5. Specify an initial estimate for the mean diameter, D_{m} Keep in mind that the mean diameter will be smaller than the OD and larger than the ID. Judgment is necessary to get started. For this problem, let’s specify D_{m} = 0.60 in. This should permit the installation into the 0.75-in-diameter hole.

Step 6. Specify an initial design stress. The charts for the design stresses for the selected material can be consulted, considering
also the service. In this problem, we should use average service. Then for the ASTM A231 steel, as shown in Figure 19_11, a nominal design stress would be 130 000 psi. This is strictly an estimate based on the strength of the material. The process includes a check on stress later.

Step 7. Compute the trial wire diameter by solving Equation (19_4) \tau=\frac{8 K F D_{m}}{\pi D_{w}^{3}}=\frac{8 K F C}{\pi D_{w}^{2}} for D_{W}. Notice that everything else in the equation is known except the Wahl factor, K, because it depends on the wire diameter itself. But K varies only little over the normal range of spring indexes, C. From Figure 19_14, note that K = 1.2 is a nominal value. This, too, will be checked later. With the assumed value of K, some simplification can be done: D_{w}=\left[\frac{8 K F_{o} D_{m}}{\pi \tau_{d}}\right]^{1 / 3}=\left[\frac{(8)(1.2)\left(F_{o}\right)\left(D_{m}\right)}{(\pi)\left(\tau_{d}\right)}\right]^{1 / 3} Combining constants gives D_{w}=\left[\frac{8 K F_{o} D_{m}}{\pi \tau_{d}}\right]^{1 / 3}=\left[\frac{(3.06)\left(F_{o}\right)\left(D_{m}\right)}{\left(\tau_{d}\right)}\right]^{1 / 3}                                 (19_7)

For this problem, D_{w}=\left[\frac{(3.06)\left(F_{o}\right)\left(D_{m}\right)}{\tau_{d}}\right]^{1 / 3}=\left[\frac{(3.06)(12)(0.6)}{130000}\right]^{0.333}

Step 8. Select a standard wire diameter from the tables, and then determine the design stress and the maximum allowable stress for the material at that diameter. The design stress will normally be for average service, unless high cycling rates or shock indicate that severe service is warranted. The light service curve should be used with care because it is
very near to the yield strength. In fact, we will use the light service curve as an estimate of the maximum allowable stress. For this problem, the next larger standard wire size is 0.0625 in, no. 16 on the U.S. Steel Wire Gage chart. For this size, the curves in Figure 19_11 for ASTM A231 steel wire show the design stress to be approximately 145 000 psi for average service, and the maximum allowable stress to be 170 000 psi from the light service curve.

Step 9. Compute the actual values of C and K, the spring index and the Wahl factor: C=\frac{D_{m}}{D_{w}}=\frac{0.60}{0.0625}=9.60

Step 10. Compute the actual expected stress due to the operating force, f_{0} , from 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} D_{m}}{\pi D_{w}^{3}}=\frac{(8)(1.15)(12.0)(0.60)}{(\pi)(0.0625)^{3}}=86450 \mathrm{psi} Comparing this with the design stress of 145 000 psi, we see that it is safe.

Step 11. Compute the number of active coils required to give the proper deflection characteristics for the spring. Using 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}} and solving for N_{a} we have f=\frac{8 F C^{3} N_{a}}{G D_{w}}

N_{a}=\frac{f G D_{w}}{8 F C^{3}}=\frac{G D_{w}}{8 k C^{3}} \quad(\text { Note } : F / f=k \text {, the spring rate. })                                     (19_8)

Then, for this problem, N_{a}=\frac{G D_{w}}{8 k C^{3}}=\frac{(11200000)(0.0625)}{(8)(8.0)(9.60)^{3}}=12.36 \text { coils } Notice that k = 8.0 lb/in is the spring rate. Do not confuse this with K. the Wahl factor.

Step 12. Compute the solid length,L_{s} ; the force on the spring at solid length, F_{s}; and the stress in the spring at solid length, \tau _{s} . This computation will give the maximum stress that the spring will receive. The solid length occurs when all of the coils are touching, but recall that there are two inactive coils for springs with squared and ground ends. Thus, L_{s}=D_{w}\left(N_{a}+2\right)=0.0625(14.36)=0.898 \text { in } The force at solid length is the product of the spring rate times the deflection to solid length\left(L_{f}-L_{s}\right) : F_{s}=k\left(L_{f}-L_{s}\right)=(8.0 \mathrm{lb} / \mathrm{in})(2.75-0.898) \text { in }=14.8 \mathrm{lb} Because the stress in the spring is directly proportional to the force, a simple method of computing the solid length stress is \tau_{s}=\left(\tau_{o}\right)\left(F_{s} / F_{o}\right)=(86450 \mathrm{psi})(14.8 / 12.0)=106750 \mathrm{psi} When this value is compared with the maximum allowable stress of 170 000 psi, we see that it is safe, and the spring will not yield when compres.sed to solid length.
Step 13. Complete the computations of geometric features, and compare them with space and operational limitations: O D=D_{m}+D_{w}=0.60+0.0625=0.663 \text { in }

I D=D_{m}-D_{w}=0.60-0.0625=0.538 \mathrm{in} These dimensions are .satisfactory for installation in a hole having a diameter of 0.75 in.
Step 14. The tendency to buckle is checked, along with the coil clearance. This procedure completes the design of one satisfactory spring for this application. It may be desirable to make other trials to attempt to find a more nearly optimum spring.