The stiffness of a helical compression spring (see Figure 19.12) can be calculated from k = d^4G/8D³n where d is the wire diameter, G is the modulus of rigidity, D is the coil diameter, n is the number of coils, and k is the spring stiffness. The mean and standard deviation for each variable are

Question

The stiffness of a helical compression spring (see Figure 19.12) can be calculated from

[latex]k=\frac{d^{4} G}{8 D^{3} n}[/latex]

where d is the wire diameter, G is the modulus of rigidity, D is the coil diameter, n is the number of coils, and k is the spring stiffness.

The mean and standard deviation for each variable are known:

[latex]\mu_{d}=2.34 mm , \mu_{D}=16.71 mm , \mu_{G}=79.29 imes 10^{3} N / mm ^{2}, \mu_{n}=14 ext { coils }[/latex]

[latex]\sigma_{d}=0.010 mm , \sigma_{D}=0.097 mm , \sigma_{G}=1.585 imes 10^{3} N / mm ^{2}, \sigma_{n}=0.0833 ext { coils. }[/latex]

Calculate the sure-fit extreme tolerance limits and the statistical basic normal tolerance limits for the spring stiffness. Assume ± 3σ natural tolerance limits: i.e. 99.73% of measurements lie within ± 3 standard deviations.

Accepted Answer

Substituting values into the equation for k to give the average value of the spring constant:

[latex]\bar{k}=\frac{2.34^{4} imes 79.29 imes 10^{3}}{8 imes 16.71^{3} imes 14}=4.549 N / mm[/latex]

[latex]\sigma_{d}=0.01 mm[/latex]

[latex]\Delta d=6 imes \sigma_{d}=0.06 mm[/latex]

[latex]\sigma_{G}=1.585 imes 10^{3}[/latex]

[latex]\Delta G=6 imes \sigma_{G}=9510 N / mm ^{2}[/latex]

[latex]\sigma_{D}=0.097[/latex]

[latex]\Delta D=6 imes \sigma_{D}=0.582 mm ext {, }[/latex]

[latex]\sigma_{n}=0.0833 ext { coils }[/latex]

[latex]\Delta n=6 imes \sigma_{n}=0.5.[/latex]

[latex]\frac{\partial k}{\partial d}=\frac{4 d^{3} G}{8 D^{3} n}=\frac{2.34^{3} imes 79.29 imes 10^{3}}{2 imes 16.71^{3} imes 14}=7.776.[/latex]

[latex]\frac{\partial k}{\partial G}=\frac{d^{4}}{8 D^{3} n}=5.7374 imes 10^{-5}.[/latex]

[latex]\frac{\partial k}{\partial D}=-\frac{3 d^{4} G}{8 D^{4} n}=-8.1673 imes 10^{-1}.[/latex]

[latex]\frac{\partial k}{\partial n}=-\frac{d^{4} G}{8 D^{3} n^{2}}=-3.2494 imes 10^{-1}[/latex]

[latex]\Delta k_{ ext {sure-fit }}=\left|\frac{\partial k}{\partial d} ight| \Delta d+\left|\frac{\partial k}{\partial G} ight| \Delta G+\left|\frac{\partial k}{\partial D} ight| \Delta D+\left|\frac{\partial k}{\partial n} ight| \Delta n[/latex]

[latex]=7.776 imes 0.06+5.7374 imes 10^{-5} imes 9510+0.81673 imes 0.582+0.32494 imes 0.5[/latex]

= 0.46656 + 0.5456 + 0.4753 + 0.16247 = 1.65 N/mm.

This figure ([latex]\Delta k_{ ext {sure-fit }}[/latex]) gives the overall worst-case variability of the spring stiffness (i.e. the total tolerance). The bilateral tolerance can be obtained by dividing Δk by 2 (i.e. 1.65/2 = 0.825).

So the spring stiffness can be stated as k = 4.549 ± 0.825 N/mm.

Calculation of the basic normal variability is

[latex]\Delta k_{ ext {basic normal }}^{2}=\left|\frac{\partial k}{\partial d} ight|^{2} \Delta d^{2}+\left|\frac{\partial k}{\partial G} ight|^{2} \Delta G^{2}+\left|\frac{\partial k}{\partial D} ight|^{2} \Delta D^{2}+\left|\frac{\partial k}{\partial n} ight|^{2} \Delta n^{2}[/latex]

= 0.21768 + 0.297708 + 0.22594 + 0.026397 = 0.7677.

[latex]\Delta k_{ ext {basic normal }}=0.8762 N / mm.[/latex]

The overall variability in the spring stiffness is calculated as 0.8762 N/mm using the basic normal method. The bilateral tolerance can be calculated by dividing this value in 2, i.e. the spring stiffness k = 4.549 ± 0.4381 N/mm.

The basic normal variation model implies a narrower bandwidth of variation k = 4.549 ± 0.438 N/mm versus k = 4.549 ± 0.825 N/mm for a sure-fit. This is more attractive from a sales perspective and accounts for 99.73% (i.e. nearly all) of the items.

If we wanted to decrease the variation in k, we would need to tighten the individual tolerances on d, G, D, and n with particular attention to d and D, as these are controllable and influential on the overall variability.

(The standard deviation of the spring stiffness can be calculated, if required, by [latex]\sigma_{k}=\Delta k / 6=0.146[/latex]).

Question 19.7: The stiffness of a helical compression spring (see Figure 19...

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