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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clc
+//Example 14.11
+//Outer diameter of clutch
+
+//------------------------------------------------------------------------------
+//Given Data:
+// number of plates
+n1=5
+n2=4
+// if n is total number of surfces
+n=n1+n2-1
+
+//Total torqe transmitting capacity
+Tt=16// Nm
+//Permissible inner diameter
+Di=0.05// m
+//coefficient of friction
+f=0.1
+//average pressure
+p=350000 //N/(m^2)
+
+//------------------------------------------------------------------------------
+//Torque per pair of surfaces
+T=Tt/n
+
+//T=F*f*((Do+Di)/4)
+//T=((%pi/4)*((Do^2)-(Di^2))*p)*f*((Do+Di)/4)
+//To solve above equation for Do, it has to brought to a polynomial equation form in Do
+//((%pi*p*f)*(Do^3))+((%pi*p*f*Di)*(Do^2))-((%pi*p*f*(Di^2))*Do)-((%pi*p*f*(Di^3))+(16*T))=0
+x=poly([-((%pi*p*f*(Di^3))+(16*T)) -(%pi*p*f*(Di^2)) (%pi*p*f*Di) (%pi*p*f)],'Do','c')
+y=roots(x)
+//y will contain all roots of the polynomial, the first of which is the acceptable one
+Do=y(1)
+
+//Axial force F
+F=T/(f*((Do+Di)/4))
+
+Do=round(Do*(10^3))
+F=round(F)
+//Actual pressure
+p=F/((%pi/4)*(((Do*(10^-3))^2)-(Di^2)))
+
+//------------------------------------------------------------------------------
+//Printing result file to .txt
+res11=mopen(TMPDIR+'11_outer_diameter_of_clutch.txt','wt')
+mfprintf(res11,"Necessary outer diameter of disks is %0.2f mm\n",Do)
+mfprintf(res11,"Necessary axial force is %0.2f N\n",F)
+mfprintf(res11,"Actual contact pressure is %0.2f kN/m^2",p*(10^-3))
+mclose(res11)
+editor(TMPDIR+'11_outer_diameter_of_clutch.txt')
+
+//------------------------------------------------------------------------------
+//-----------------------------End of program----------------------------------- \ No newline at end of file