Question

A single plate clutch using both contact planes is used to deliver a torque of 90 Nm. The axial contact pressure on the clutch is 84 kPa, the coefficient of friction is 0.29 and the outer diameter is 1.26 times the inner diameter. Calculate the inner and outer diameters of the clutch required, as well as the required axial force. 

Answer 1

T=90Nm

P=84kPa=84*10³Nm²

D_o=1.26D_i

$\mu=0.29$

$T=\int_{\frac{D_i}{2}}^{\frac{D_o}{2}}2π\mu Pr^{2}dr$

$=2π\mu P[\frac{r^{3}}{3}]_{\frac{D_i}{2}}^{\frac{D_o}{2}}$

$=\frac{2}{3}π\mu P[\frac{D_o}{2}-\frac{D_i}{2}]$

$=\frac{1}{3}π\mu P(D_o^{3}-D_i^{3})$

Substitute the values we have to get

$90=\frac{1}{3}π*0.29*84*10^{3}(1.26D_i^{3}-D_i^{3})\newline 90=25509.73(0.26D_i)\newline D_i^{3}=\frac{90}{25509.73}=3.5281*10^{-3}\newline D_i=\sqrt[3]{3.5281*10^{-3}}\newline D_i=0.1522m\newline D_o=1.26*D_i=1.26*0.1522m\newline D_o=0.1918m\newline Axial\ force=\frac{π}{4}P(D_o^{2}-D_i^{2})\newline =\frac{π}{4}*84*10^{3}(0.1918^{2}-0.1522^{2})\newline =898.7N$