Question 19.7: The stiffness of a helical compression spring (see Figure 19...
The stiffness of a helical compression spring (see Figure 19.12) can be calculated from
k=\frac{d^{4} G}{8 D^{3} n}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:
\mu_{d}=2.34 mm , \mu_{D}=16.71 mm , \mu_{G}=79.29 \times 10^{3} N / mm ^{2}, \mu_{n}=14 \text { coils }\sigma_{d}=0.010 mm , \sigma_{D}=0.097 mm , \sigma_{G}=1.585 \times 10^{3} N / mm ^{2}, \sigma_{n}=0.0833 \text { coils. }
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.
Substituting values into the equation for k to give the average value of the spring constant:
\bar{k}=\frac{2.34^{4} \times 79.29 \times 10^{3}}{8 \times 16.71^{3} \times 14}=4.549 N / mm\sigma_{d}=0.01 mm
\Delta d=6 \times \sigma_{d}=0.06 mm
\sigma_{G}=1.585 \times 10^{3}
\Delta G=6 \times \sigma_{G}=9510 N / mm ^{2}
\sigma_{D}=0.097
\Delta D=6 \times \sigma_{D}=0.582 mm \text {, }
\sigma_{n}=0.0833 \text { coils }
\Delta n=6 \times \sigma_{n}=0.5.
\frac{\partial k}{\partial d}=\frac{4 d^{3} G}{8 D^{3} n}=\frac{2.34^{3} \times 79.29 \times 10^{3}}{2 \times 16.71^{3} \times 14}=7.776.
\frac{\partial k}{\partial G}=\frac{d^{4}}{8 D^{3} n}=5.7374 \times 10^{-5}.
\frac{\partial k}{\partial D}=-\frac{3 d^{4} G}{8 D^{4} n}=-8.1673 \times 10^{-1}.
\frac{\partial k}{\partial n}=-\frac{d^{4} G}{8 D^{3} n^{2}}=-3.2494 \times 10^{-1}
\Delta k_{\text {sure-fit }}=\left|\frac{\partial k}{\partial d}\right| \Delta d+\left|\frac{\partial k}{\partial G}\right| \Delta G+\left|\frac{\partial k}{\partial D}\right| \Delta D+\left|\frac{\partial k}{\partial n}\right| \Delta n
=7.776 \times 0.06+5.7374 \times 10^{-5} \times 9510+0.81673 \times 0.582+0.32494 \times 0.5
= 0.46656 + 0.5456 + 0.4753 + 0.16247 = 1.65 N/mm.
This figure (\Delta k_{\text {sure-fit }}) 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
\Delta k_{\text {basic normal }}^{2}=\left|\frac{\partial k}{\partial d}\right|^{2} \Delta d^{2}+\left|\frac{\partial k}{\partial G}\right|^{2} \Delta G^{2}+\left|\frac{\partial k}{\partial D}\right|^{2} \Delta D^{2}+\left|\frac{\partial k}{\partial n}\right|^{2} \Delta n^{2}= 0.21768 + 0.297708 + 0.22594 + 0.026397 = 0.7677.
\Delta k_{\text {basic normal }}=0.8762 N / mm.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 \sigma_{k}=\Delta k / 6=0.146).