The torque capacity, T, for a multiple disc clutch can be calculated using the relationship T =2/3 µ FaN ( r³ o - r³/ r² o - r²) where µ is the coefficient of friction, Fa is the axial clamping force, N is the number of disc faces, ro is the outer radius of clutch ring, and ri is the inner radius

Question

The torque capacity, T, for a multiple disc clutch can be calculated using the relationship

[latex]T=\frac{2}{3} \mu F_{a} N\left(\frac{r_{o}^{3}-r_{i}^{3}}{r_{o}^{2}-r_{i}^{2}}\right)[/latex]

where µ is the coefficient of friction, [latex]F_a[/latex] is the axial clamping force, N is the number of disc faces, [latex]r_o[/latex] is the outer radius of clutch ring, and [latex]r_i[/latex] is the inner radius of clutch ring.

Determine the sure-fit and basic normal probable limits for the torque capacity if

[latex]\mu=0.3 \pm 0.03,[/latex]

[latex]F_{a}=4000 \pm 200 N ,[/latex]

N = 12,

[latex]r_{o}=60 \pm 0.5 mm[/latex]

[latex]r_{i}=30 \pm 0.5 mm.[/latex]

Which design parameter should be reduced to create a maximum reduction in the variability of the torque capacity?

Accepted Answer

[latex]T_{a v}=\frac{2}{3} imes 0.3 imes 4000 imes 12 imes\left(\frac{0.06^{3}-0.03^{3}}{0.06^{2}-0.03^{2}} ight)=672 N m[/latex]

For sure-fit variability,

[latex]\delta y \approx \sum_{j=1}^{n} \frac{\partial f}{\partial x_{j}} \delta x_{j}[/latex]

[latex]\Delta T_{ ext {sure-fit }} \approx\left|\frac{\partial T}{\partial F_{a}} ight| \Delta F_{a}+\left|\frac{\partial T}{\partial \mu} ight| \Delta \mu+\left|\frac{\partial T}{\partial r_{o}} ight| \Delta r_{o}+\left|\frac{\partial T}{\partial r_{i}} ight| \Delta r_{i}[/latex]

[latex]\frac{\partial T}{\partial F_{a}}=\frac{2}{3} \mu N \frac{r_{o}^{3}-r_{i}^{3}}{r_{o}^{2}-r_{i}^{2}}=\frac{2}{3} imes 0.3 imes 12 imes \frac{0.06^{3}-0.03^{3}}{0.06^{2}-0.03^{2}}=0.168[/latex]

[latex]\frac{\partial T}{\partial \mu}=\frac{2}{3} F_{a} N \frac{r_{o}^{3}-r_{i}^{3}}{r_{o}^{2}-r_{i}^{2}}=\frac{2}{3} imes 4000 imes 12 imes \frac{0.06^{3}-0.03^{3}}{0.06^{2}-0.03^{2}}=2240[/latex]

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

[latex]\frac{\partial T}{\partial r_{o}}=\frac{2}{3} \mu F_{a} N \frac{\left(r_{o}^{2}-r_{i}^{2} ight) 3 r_{o}^{2}-\left(r_{o}^{3}-r_{i}^{3} ight) 2 r_{o}}{\left(r_{o}^{2}-r_{i}^{2} ight)^{2}}[/latex]

[latex]=\frac{2}{3} imes 0.3 imes 4000 imes 12 imes \frac{2.7 imes 10^{-3} imes 3 imes 0.06^{2}-1.89 imes 10^{-4} imes 2 imes 0.06}{\left(0.06^{2}-0.03^{2} ight)^{2}}[/latex]

= 0.8889 × 9600 = 8533

[latex]\frac{\partial T}{\partial r_{i}}=\frac{2}{3} \mu F_{a} N \frac{-\left(r_{o}^{2}-r_{i}^{2} ight) 3 r_{i}^{2}+\left(r_{o}^{3}-r_{i}^{3} ight) 2 r_{i}}{\left(r_{o}^{2}-r_{i}^{2} ight)^{2}}[/latex]

[latex]=\frac{2}{3} imes 0.3 imes 4000 imes 12 imes \frac{2.7 imes 10^{-3} imes 3 imes 0.03^{2}-1.89 imes 10^{-4} imes 2 imes 0.03}{\left(0.06^{2}-0.03^{2} ight)^{2}}[/latex]

= 0.5556 × 9600 = 5333

[latex]\Delta F_{a}=2 imes 200=400 N[/latex]

[latex]\Delta \mu=2 imes 0.3=0.6 N[/latex]

[latex]\Delta r_{o}=2 imes 0.0005=0.001 m[/latex]

[latex]\Delta r_{i}=2 imes 0.0005=0.001 m[/latex]

[latex]\Delta T_{ ext {sure-fit }}=(2240 imes 0.06)+(0.168 imes 400)+(0.8889 imes 0.001 imes 9600) +(0.5556 imes 0.001 imes 9600)[/latex]
=134.4+67.2+8.533+5.333=215.5 N m

[latex]T_{ ext {sure-fit }}=T_{a v} \pm\left(\frac{\Delta T_{ ext {sure-fit }}}{2} ight)=672 \pm 108 N m[/latex]

Basic normal probable limits are [latex]\sigma_{y}^{2} \approx \sum_{j=1}^{n}\left(\frac{\partial f}{\partial x_{j}} ight)^{2} \sigma_{x_{j}}^{2}[/latex].

Assuming natural tolerance limits,

[latex]\Delta T_{ ext {basic normal }}^{2} \approx\left|\frac{\partial T}{\partial F_{a}} ight|^{2} \Delta F_{a}^{2}+\left|\frac{\partial T}{\partial \mu} ight|^{2} \Delta \mu^{2}+\left|\frac{\partial T}{\partial r_{o}} ight|^{2} \Delta r_{o}^{2}+\left|\frac{\partial T}{\partial r_{i}} ight|^{2} \Delta r_{i}^{2}[/latex]

[latex]=134.4^{2}+67.2^{2}+8.533^{2}+5.333^{2}=22,680[/latex]

[latex]\Delta T_{ ext {basic normal }}=150.6 N m[/latex]

[latex]T_{ ext {basic normal }}=T_{a v} \pm\left(\frac{\Delta T_{ ext {basic normal }}}{2} ight)=672 \pm 75 N m[/latex]

The coefficient of friction contributes the most to the variability. This would require more stringent manufacture control.

Question 19.9: The torque capacity, T, for a multiple disc clutch can be ca...

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