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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// find..
+clc
+//solution
+//given
+P=35000//W
+N=240//rpm
+L=0.350//mm
+u=0.25
+t=10//mm
+n=6
+q=%pi
+fts=80//N/mm^2
+ftk=80//N/mm^2
+ts=50//N/mm^2
+tk=50//N/mm^2
+f=2.5//N/mm^2
+ft=4.5//N/mm^2
+fb=15//N/mm^2
+rho=7200//kg/m^3
+//fb*10^6=rho*v^2
+v=sqrt(ft*10^6/rho)//m/s
+printf("the speed is,%f m/s\n",v)
+D=(v*60)/(%pi*N)//m
+printf("the dia of pulley is,%f m\n",D)
+//let b e width of belt
+//let T1 and T2 be tension on tight and slag side
+//P=(T1-T2)*v
+//T1-T2=1400//N
+//log(T1/T2)=u*q=0.25*3.14=0.3415
+//T2=T1/2.195//N
+T1=2572//N
+T2=1172//N
+rho1=1000//kg/m^3
+//A=b*t=10*b/10^6//m^2
+//let b is width
+//m=A*rho1=b*t*rho=0.01*b//kg/m
+//Tc=m*v^2=6.25*b
+//T=f*b*t=25*b//N
+//T1=T-Tc
+//T1=25*b-6.25b
+//b=T1/18.75//mm
+printf("the width of belt si,%f mm\n",T1/18.75)
+printf("the standard width of belt is b=140mm\n")
+b=140//mm
+Tc=6.25*b//N
+//let d eb dia of shaft
+T3=(P*60)/(2*%pi*N)//N-mm
+M=(T1+T2+2*Tc)*L
+Te=sqrt(M^2 + T3^2)
+//d=((Te*16*1000)/(%pi*ts))^(1/3)//mm
+printf("the dia of shaft is,%f mm\n",((Te*16*1000)/(%pi*ts))^(1/3))
+printf("the standard dia of shaft is d=65mm\n")
+d=65//mm
+//corres to d=65mm,
+width =20//mm
+thickness =12//mm
+printf("correponding to d=65mm,thickness and width is,%f mm\n,%f mm\n",thickness,width)
+//let l be length of key
+//T3*1000=l*20*tk*d/2
+//l=T3/(32500)//mm
+printf("the length of key is,%f mm\n",T3/(32500))
+printf("length should be atleast l=102 mm\n")
+printf("therfore length is 102 mm\n")
+l=102//mm
+//let b1 minor axis,a1 major axis
+M=2*T3/n*1000//N-mm
+//a1=2b1
+//Z=(%pi/32)*b1*a1^2=(%pi)/8*b1^3=0.2=393*b1^3
+//fb=M/Z
+b1=(M/(0.393*fb))^(1/3)//mm
+printf("the minor and moajor axis is,%f mm\n,%f mm",b1,2*b1)
+