For uniform wear "p\\cdot r=C=const" . Since
"p\\cdot r_2=0.1\\cdot120=12N\/mm"
Axial force required to engage the clutch
"W=2\\pi C(r_1-r_2)"
The torque transmitted
"T=4\\cdot\\mu W\\frac{r_1+r_2}{2}=4\\cdot \\mu\\cdot 2\\pi C(r_1-r_2)\\cdot\\frac{r_1+r_2}{2}=4\\cdot\\mu\\cdot \\pi C(r_1-r_2)\\cdot (r_1+r_2)"
Power transmitted
"P=T\\cdot \\omega=4\\cdot\\mu\\cdot \\pi C(r_1-r_2)\\cdot (r_1+r_2)\\cdot 2\\pi \\nu" "\\to"
"r_1=\\sqrt{\\frac{P}{8\\mu\\pi^2C\\nu}+r_2^2}=\\sqrt{\\frac{25000}{8\\cdot0.3\\cdot3.14^2\\cdot12000\\cdot26.25}+0.12^2}=0.158m=158mm" Answer.
Comments
Leave a comment