For a standard series “A” Belleville spring, the spring rate or stiffness can be determined by the equation k = 4E/(1 - µ²) × t³/K1D²e where E is Young’s modulus of the washer material, t is the washer thickness, m is Poisson’s ration,K1 is a dimensionless constant, and De is the external diameter

Question

For a standard series “A” Belleville spring, the spring rate or stiffness can be determined by the equation

[latex]k=\frac{4 E}{\left(1-\mu^{2} ight)} imes \frac{t^{3}}{K_{1} D_{e}^{2}}[/latex]

where E is Young’s modulus of the washer material, t is the washer thickness, µ is Poisson’s ration, [latex]K_1[/latex] is a dimensionless constant, and [latex]D_e[/latex] is the external diameter.

Determine the sure-fit and probable limits for the spring stiffness stating any assumptions made if

[latex]t=2.22 \pm 0.03 mm , D_{e}=40.00 \pm 0.08 mm , \mu=0.30 \pm 0.003,[/latex]

[latex]E=207 imes 10^{9} \pm 2 imes 10^{9} N / m ^{2}, K_{1}=0.69[/latex]

Comment on which tolerance could be altered to best reduce the overall variability.

Accepted Answer

[latex]k=\frac{4 E}{\left(1-\mu^{2} ight)} imes \frac{t^{3}}{K_{1} D_{e}^{2}}.[/latex]

[latex]K_{1}= ext { constant }.[/latex]

Assuming quantities are subject to variability and are uncorrelated and random,

[latex]\Delta k_{ ext {sure-fit }}=\left|\frac{\partial k}{\partial E} ight| \Delta E+\left|\frac{\partial k}{\partial \mu} ight| \Delta \mu+\left|\frac{\partial k}{\partial t} ight| \Delta t+\left|\frac{\partial k}{\partial D_{e}} ight| \Delta D_{e}[/latex].

[latex]\Delta k_{ ext {basic normal }}^{2}=\left|\frac{\partial k}{\partial E} ight|^{2} \Delta E^{2}+\left|\frac{\partial k}{\partial \mu} ight|^{2} \Delta \mu^{2}+\left|\frac{\partial k}{\partial t} ight|^{2} \Delta t^{2}+\left|\frac{\partial k}{\partial D_{e}} ight|^{2} \Delta D_{e}^{2} ext {. }[/latex]

[latex]\frac{\partial k}{\partial E}=\frac{4 t^{3}}{\left(1-\mu^{2} ight) K_{1} D_{e}^{2}}=\frac{4 imes\left(2.22 imes 10^{-3} ight)^{3}}{\left(1-0.3^{2} ight) imes 0.69 imes 0.04^{2}}=4.356 imes 10^{-5}.[/latex]

[latex]\frac{\partial k}{\partial \mu}=-\frac{2 \mu}{\left(1-\mu^{2} ight)^{2}} imes \frac{4 E t^{3}}{K_{1} D_{e}^{2}}=-\frac{2 imes 0.3}{(1-0.09)^{2}} imes \frac{4 imes 207 imes 10^{9} imes\left(2.22 imes 10^{-3} ight)^{3}}{0.69 imes 0.04^{2}}[/latex]

[latex]=-5.946 imes 10^{6}.[/latex]

Quotient rule, [latex]\left(\frac{u}{v} ight)^{\prime}=\frac{u v^{\prime}-v u^{\prime}}{v^{2}}[/latex]

[latex]\frac{\partial k}{\partial t}=\frac{12 E t^{2}}{\left(1-\mu^{2} ight) K_{1} D_{e}^{2}}=\frac{12 imes 207 imes 10^{9} imes\left(2.22 imes 10^{-3} ight)^{2}}{(1-0.09) imes 0.69 imes 0.04^{2}}=1.219 imes 10^{10}[/latex]

[latex]\frac{\partial k}{\partial D_{e}}=-\frac{8 E t^{3}}{\left(1-\mu^{2} ight) K_{1} D_{e}^{3}}=-\frac{8 imes 207 imes 10^{9} imes(0.00222)^{3}}{(1-0.09) imes 0.69 imes(0.04)^{3}}=-4.509 imes 10^{8}.[/latex]

Assume natural tolerance limits ± 3σ,

[latex]\Delta E=2 imes 2 imes 10^{9}=4 imes 10^{9}.[/latex]

[latex]\Delta \mu=2 imes 0.003=0.006.[/latex]

[latex]\Delta t=2 imes 0.03 imes 10^{-3}=6 imes 10^{-5} .[/latex]

[latex]\Delta D_{e}=2 imes 0.08 imes 10^{-3}=1.6 imes 10^{-4}.[/latex]

[latex]\Delta k_{ ext {sure-fit }}=(4.356 imes 10-5-5 imes 4 imes 109)+(5.946 imes 106 imes 0.006) +\left(1.219 imes 10^{10} imes 6 imes 10^{-5} ight)+\left(4.509 imes 10^{8} imes 1.6 imes 10^{-4} ight)[/latex]
=174,240+35,676+731,400+72,144=1,013,460

[latex]\bar{k}=\frac{4 imes 207 imes 10^{9}}{1-0.3^{2}} imes \frac{\left(2.22 imes 10^{-3} ight)^{3}}{0.69 imes 0.04^{2}}=9,017,347.3 N / m.[/latex]

[latex]k_{ ext {sure-fit }}=9,017,000 \pm 506,700 N / m.[/latex]

[latex]\Delta k_{ ext {basic normal }}^{2}=174,240^{2}+35,676^{2}+731,400^{2}+72,144^{2}=5.718 imes 10^{11}[/latex]

[latex]\Delta k_{ ext {basic normal }}=756,200.[/latex]

[latex]k_{ ext {basic normal }}=9,017,000 \pm 378,100 N / m.[/latex]

To reduce the variability, tighten the tolerances on t as this has the highest sensitivity coefficient.

Question 19.10: For a standard series “A” Belleville spring, the spring rate...

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