Question 19.9: The torque capacity, T, for a multiple disc clutch can be ca...
The torque capacity, T, for a multiple disc clutch can be calculated using the relationship
T=\frac{2}{3} \mu F_{a} N\left(\frac{r_{o}^{3}-r_{i}^{3}}{r_{o}^{2}-r_{i}^{2}}\right)where µ is the coefficient of friction, F_a is the axial clamping force, N is the number of disc faces, r_o is the outer radius of clutch ring, and r_i is the inner radius of clutch ring.
Determine the sure-fit and basic normal probable limits for the torque capacity if
\mu=0.3 \pm 0.03,F_{a}=4000 \pm 200 N ,
N = 12,
r_{o}=60 \pm 0.5 mmr_{i}=30 \pm 0.5 mm.
Which design parameter should be reduced to create a maximum reduction in the variability of the torque capacity?
For sure-fit variability,
\delta y \approx \sum_{j=1}^{n} \frac{\partial f}{\partial x_{j}} \delta x_{j}\Delta T_{\text {sure-fit }} \approx\left|\frac{\partial T}{\partial F_{a}}\right| \Delta F_{a}+\left|\frac{\partial T}{\partial \mu}\right| \Delta \mu+\left|\frac{\partial T}{\partial r_{o}}\right| \Delta r_{o}+\left|\frac{\partial T}{\partial r_{i}}\right| \Delta r_{i}
\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} \times 0.3 \times 12 \times \frac{0.06^{3}-0.03^{3}}{0.06^{2}-0.03^{2}}=0.168
\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} \times 4000 \times 12 \times \frac{0.06^{3}-0.03^{3}}{0.06^{2}-0.03^{2}}=2240
Quotient rule, \left(\frac{u}{v}\right)^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^{2}}.
\frac{\partial T}{\partial r_{o}}=\frac{2}{3} \mu F_{a} N \frac{\left(r_{o}^{2}-r_{i}^{2}\right) 3 r_{o}^{2}-\left(r_{o}^{3}-r_{i}^{3}\right) 2 r_{o}}{\left(r_{o}^{2}-r_{i}^{2}\right)^{2}}
=\frac{2}{3} \times 0.3 \times 4000 \times 12 \times \frac{2.7 \times 10^{-3} \times 3 \times 0.06^{2}-1.89 \times 10^{-4} \times 2 \times 0.06}{\left(0.06^{2}-0.03^{2}\right)^{2}}
= 0.8889 × 9600 = 8533
\frac{\partial T}{\partial r_{i}}=\frac{2}{3} \mu F_{a} N \frac{-\left(r_{o}^{2}-r_{i}^{2}\right) 3 r_{i}^{2}+\left(r_{o}^{3}-r_{i}^{3}\right) 2 r_{i}}{\left(r_{o}^{2}-r_{i}^{2}\right)^{2}}=\frac{2}{3} \times 0.3 \times 4000 \times 12 \times \frac{2.7 \times 10^{-3} \times 3 \times 0.03^{2}-1.89 \times 10^{-4} \times 2 \times 0.03}{\left(0.06^{2}-0.03^{2}\right)^{2}}
= 0.5556 × 9600 = 5333
\Delta F_{a}=2 \times 200=400 N\Delta \mu=2 \times 0.3=0.6 N
\Delta r_{o}=2 \times 0.0005=0.001 m
\Delta r_{i}=2 \times 0.0005=0.001 m
\Delta T_{\text {sure-fit }}=(2240 \times 0.06)+(0.168 \times 400)+(0.8889 \times 0.001 \times 9600) +(0.5556 \times 0.001 \times 9600)
=134.4+67.2+8.533+5.333=215.5 N m
Basic normal probable limits are \sigma_{y}^{2} \approx \sum_{j=1}^{n}\left(\frac{\partial f}{\partial x_{j}}\right)^{2} \sigma_{x_{j}}^{2}.
Assuming natural tolerance limits,
\Delta T_{\text {basic normal }}^{2} \approx\left|\frac{\partial T}{\partial F_{a}}\right|^{2} \Delta F_{a}^{2}+\left|\frac{\partial T}{\partial \mu}\right|^{2} \Delta \mu^{2}+\left|\frac{\partial T}{\partial r_{o}}\right|^{2} \Delta r_{o}^{2}+\left|\frac{\partial T}{\partial r_{i}}\right|^{2} \Delta r_{i}^{2}=134.4^{2}+67.2^{2}+8.533^{2}+5.333^{2}=22,680
\Delta T_{\text {basic normal }}=150.6 N m
T_{\text {basic normal }}=T_{a v} \pm\left(\frac{\Delta T_{\text {basic normal }}}{2}\right)=672 \pm 75 N m
The coefficient of friction contributes the most to the variability. This would require more stringent manufacture control.