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+//design wire rope
+clc
+//soltuion
+//given
+W=55000//N
+depth=300//m
+v=500//m/min
+t=10//s
+//ref T20.6,we choose rope type 6*19
+//,tab; 20.11,Fs =7,for depth 300 to 600m,design load is calculated by taking 2 to 2.5 times factor of safety fiven is table
+//ref table 20.11
+Designload=15*55*1000//N
+//ref table 20.6,tesnile strength of 6*19 is=595*d^2
+//595d^2=designload
+//d=sqrt(Desingload/595)//mm
+printf("the dia of rope is,%f mm\n",sqrt(Designload/595))
+printf("the dia of rope is,say 38mm\n")
+d=38//mm
+dw=0.063*d//ref table 20.10,dw=dia of wire
+A=0.38*d^2
+//ref table 20.6
+w=0.0363*d^2*depth//N
+//ref table 20.12
+D=100*d
+fb=84000*dw/D
+printf("bending stress acting is,%f N/mm^2\n",fb)
+Wb=fb*A//N
+printf("the bending load on rope is,%f N\n",Wb)
+a=v/(60*t)//acceleration
+g=9.81//m/s^2
+Wa=(W+w)/g*a//additonal load
+printf("additional load due to acc si,%f N\n",Wa)
+Wst=2*(W+w)
+printf("the starting load acting is,%f N\n",Wst)
+We=W+w+Wb//N
+printf("effctive load during uniform velocity is,%f N\n",We)
+Fsa=Designload/We
+printf("actual factor of safety is,%f \n",Fsa)
+printf("since factor of safety caculated above are safe,therefore wire rope of dia 38mm and 6*19 is chosen") \ No newline at end of file