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+//find dia o hnalde,xsec,dia of journal
+clc
+//solution
+//given
+l=300//mm
+L=400//mm
+x=100//mm
+P=400//N
+ft=50//N/mm^2
+t=40//N/mm62
+//let d eb dia
+M=(1-1/3)*P*l//N-mm
+//Z=(%pi/32)*d^3=0.0982*d^3
+//M=fb*Z=4.91*d^3
+d=(M/4.91)^(1/3)//N-mm
+printf("the dia of handle is,%f mm\n",d)
+//let t1 be thicnes and B eb width of lvever arm
+M1=1.25*P*L//N-mm
+//B=2t
+//Z1=(1/6)*t*B^2=0.6677*t^3
+//ft=M/Z
+//t1=(M1/(0.667*50))^(1/3)//mm
+printf("the thcikness is,%f mm\n",(M1/(0.667*50))^(1/3))
+//let D be dia of journal
+printf("the thickness of lever arm is say 20 mm\n")
+t1=20//mm
+B=2*t1//mm
+printf("the width of lever arm is,%f mm\n",B)
+Te=P*(sqrt((2*(l/3) + x)^2 + L^2 ))//N-mm
+//Te=(%pi/16)*t*D^3=7.86*D^3
+D=(Te/7.86)^(1/3)//mm
+printf("the dia met of journal is,%f mm\n",D)