Question 19.10: For a standard series “A” Belleville spring, the spring rate...
For a standard series “A” Belleville spring, the spring rate or stiffness can be determined by the equation
k=\frac{4 E}{\left(1-\mu^{2}\right)} \times \frac{t^{3}}{K_{1} D_{e}^{2}}where E is Young’s modulus of the washer material, t is the washer thickness, µ is Poisson’s ration, K_1 is a dimensionless constant, and D_e is the external diameter.
Determine the sure-fit and probable limits for the spring stiffness stating any assumptions made if
t=2.22 \pm 0.03 mm , D_{e}=40.00 \pm 0.08 mm , \mu=0.30 \pm 0.003,E=207 \times 10^{9} \pm 2 \times 10^{9} N / m ^{2}, K_{1}=0.69
Comment on which tolerance could be altered to best reduce the overall variability.
K_{1}=\text { constant }.
Assuming quantities are subject to variability and are uncorrelated and random,
\Delta k_{\text {sure-fit }}=\left|\frac{\partial k}{\partial E}\right| \Delta E+\left|\frac{\partial k}{\partial \mu}\right| \Delta \mu+\left|\frac{\partial k}{\partial t}\right| \Delta t+\left|\frac{\partial k}{\partial D_{e}}\right| \Delta D_{e}.
\Delta k_{\text {basic normal }}^{2}=\left|\frac{\partial k}{\partial E}\right|^{2} \Delta E^{2}+\left|\frac{\partial k}{\partial \mu}\right|^{2} \Delta \mu^{2}+\left|\frac{\partial k}{\partial t}\right|^{2} \Delta t^{2}+\left|\frac{\partial k}{\partial D_{e}}\right|^{2} \Delta D_{e}^{2} \text {. }
\frac{\partial k}{\partial E}=\frac{4 t^{3}}{\left(1-\mu^{2}\right) K_{1} D_{e}^{2}}=\frac{4 \times\left(2.22 \times 10^{-3}\right)^{3}}{\left(1-0.3^{2}\right) \times 0.69 \times 0.04^{2}}=4.356 \times 10^{-5}.
\frac{\partial k}{\partial \mu}=-\frac{2 \mu}{\left(1-\mu^{2}\right)^{2}} \times \frac{4 E t^{3}}{K_{1} D_{e}^{2}}=-\frac{2 \times 0.3}{(1-0.09)^{2}} \times \frac{4 \times 207 \times 10^{9} \times\left(2.22 \times 10^{-3}\right)^{3}}{0.69 \times 0.04^{2}}
=-5.946 \times 10^{6}.
Quotient rule, \left(\frac{u}{v}\right)^{\prime}=\frac{u v^{\prime}-v u^{\prime}}{v^{2}}
\frac{\partial k}{\partial t}=\frac{12 E t^{2}}{\left(1-\mu^{2}\right) K_{1} D_{e}^{2}}=\frac{12 \times 207 \times 10^{9} \times\left(2.22 \times 10^{-3}\right)^{2}}{(1-0.09) \times 0.69 \times 0.04^{2}}=1.219 \times 10^{10}
\frac{\partial k}{\partial D_{e}}=-\frac{8 E t^{3}}{\left(1-\mu^{2}\right) K_{1} D_{e}^{3}}=-\frac{8 \times 207 \times 10^{9} \times(0.00222)^{3}}{(1-0.09) \times 0.69 \times(0.04)^{3}}=-4.509 \times 10^{8}.
Assume natural tolerance limits ± 3σ,
\Delta E=2 \times 2 \times 10^{9}=4 \times 10^{9}.\Delta \mu=2 \times 0.003=0.006.
\Delta t=2 \times 0.03 \times 10^{-3}=6 \times 10^{-5} .
\Delta D_{e}=2 \times 0.08 \times 10^{-3}=1.6 \times 10^{-4}.
\Delta k_{\text {sure-fit }}=(4.356 \times 10-5-5 \times 4 \times 109)+(5.946 \times 106 \times 0.006) +\left(1.219 \times 10^{10} \times 6 \times 10^{-5}\right)+\left(4.509 \times 10^{8} \times 1.6 \times 10^{-4}\right)
=174,240+35,676+731,400+72,144=1,013,460
k_{\text {sure-fit }}=9,017,000 \pm 506,700 N / m.
\Delta k_{\text {basic normal }}^{2}=174,240^{2}+35,676^{2}+731,400^{2}+72,144^{2}=5.718 \times 10^{11}
\Delta k_{\text {basic normal }}=756,200.
k_{\text {basic normal }}=9,017,000 \pm 378,100 N / m.
To reduce the variability, tighten the tolerances on t as this has the highest sensitivity coefficient.