Question 19.EP.3: The process is similar to that of Method 1, so only the majo...

Solution Method 2Trial 1        1. The general procedure is outlined at the top of the spreadsheet. The designer selects a material, estimates the wire diameter as an initial trial, and inputs a corresponding estimate of the design stress.
2. The spreadsheet then computes a new trial wire diameter using a formula derived from the fundamental equation for shear stress in a helical compression spring. Equation (19_4). The development is described here. \tau=\frac{8 K F C}{\pi D_{w}^{2}}               (19_4)

Let F=F_{o} \text { and } \tau=\tau_{d} (the design stress). Solving for the wire diameter gives D_{w}=\sqrt{\frac{8 K F_{o} C}{\pi \tau_{d}}}              (19_9)

The values of K and C are not yet known, but a good estimate for the wire diameter can be computed if the spring index is assumed to be approximately 7.0, a reasonable value. The corresponding value for the Wahl factor is k = 1.2, from Equation (19_5). K=\frac{4 C-1}{4 C-4}+\frac{0.615}{C} Combining these assumed values with the other constants in the preceding equation gives D_{w}=\sqrt{21.4\left(F_{o}\right) /\left(\tau_{d}\right)}                                  (19-10)

This formula is programmed into the spreadsheet in the cell to the right of the one labeled D_{wt}, the computed trial wire diameter

3. The designer then enters a standard wire size and determines revised values for the design stress and the maximum allowable stress from the graphs of material properties, Figures 19_8 through 19_13.

4. The spreadsheet then computes the maximum permissible number of active coils for the spring. The logic here is that the solid length must be less than the operating length. The solid length is the product of the wire diameter and the total number of coils. For squared and ground ends, this is L_{s}=D_{w}\left(N_{a}+2\right) Note that different relationships are used for the total number of coils for springs with other end conditions. See the discussion “Number of Coils” in Section 19_3. Now letting L_{\mathrm{s}}=L_{o}as a limit and solving for the number of coils gives \left(N_{a}\right)_{\max }=\left(L_{o}-2 D_{w}\right) / D_{w}     (19-11)

This is the formula programmed into the spreadsheet cell to the right of the one labeled N_{\max }.

5. The designer now has the freedom to choose any number of active coils less than the computed maximum value. Note the effects of that decision. Choosing a small number of coils will provide more clearance between adjacent coils and will use less wire per spring. However, the stresses produced for a given load will be higher, so there is a practical limit. One approach is to try progressively fewer coils until the stress approaches the design stress. Whereas any number of coils can be produced, even fractional numbers, we suggest trying integer values for the convenience of the manufacturer.

6. After the selected number of coils has been entered, the spreadsheet can complete the remaining calculations. One additional new formula is used in this spreadsheet to compute the value of the spring index, C. It is developed from the second form of 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}} relating the deflection for the spring,f, to a corresponding applied force, F, the value of C, and other parameters that are already known. First we solve for C^{3}: f=\frac{8 F D_{m}^{3} N_{a}}{G D_{w}^{4}}=\frac{8 F C^{3} N_{a}}{G D_{w}}

C^{3}=\frac{f G D_{w}}{8 F N_{a}} Now notice that we have the force F in the denominator and the corresponding deflection f in the numerator. But the spring rate k is defined as the ratio of F/f Then we can substitute k into the denominator and solve for C: C=\left[\frac{G D_{w}}{8 k N_{a}}\right]^{1 / 3}(19-12)

This formula is programmed into the spreadsheet cell to the right of the one labeled C =.

7. Recall that C is defined as the ratio, D_{m} / D_{w}. We can now solve for the mean diameter: D_{m}=C D_{w} This is used to compute the mean diameter in the cell to the right of D_{m}=.

8. The remaining calculations use equations already developed and used earlier. Again, the designer is responsible for evaluating the suitability of the results and for performing any additional iterations to search for an optimum result.

Solution Method 2,Trial 2         Now notice that the solution obtained in Trial I and shown in Figure 19_17 is far from optimum. Its free length of 2.75 in is quite long compared with the mean diameter of 0.343 in. The buckling ratio of L_{f} / D_{m}= 8.01 indicates that the spring is long and slender. Checking Figure 19_15, we can .see that buckling is predicted.One way to work toward a more suitable geometry is to increase the wire diameter and reduce the number of coils. The net result will be a larger mean diameter, improving the buckling ratio. Figure 19_18 shows the result of several iterations, finally using D_{w}=0.0625 \text { in } (larger than the former value of 0.0475 in) and 16 active coils (down from 22 in the first trial). The buckling ratio is down to 4.99, indicating that buckling is unlikely. The stress at operating force is comfortably lower than the design stress. The other geometrical features also appear to be satisfactory.