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+// exa 7.1 Pg 195
+clc;clear;close;
+
+// Given Data
+P=20;// kW
+N=240;// rpm
+tau_s=45;// MPa
+tau_b=30;// MPa
+sigma_b=60;// MPa
+sigma_cs=2*tau_s;// MPa
+tau_ci=15;// MPa
+//Tmax=1.25*Tm
+mu=0.15;// coefficient of friction
+
+//SHAFT DIAMETER
+// P= 2*%pi*N*Tm/60/1000 
+Tm=P/(2*%pi*N/60/1000);// N.m
+Tmax=1.25*Tm;// N.m
+// %pi*d**3*tau_s/16= Tmax
+d=(Tmax/(%pi*tau_s/16)*1000)**(1/3);// mm
+printf('shaft diameter = %.2f mm. Use d = 50 mm.',d)
+d=50;// mm
+
+// HUB DIAMETER
+// Tmax=%pi/16*((d1**4-d**4)/d1)*tau_h
+tau_h=tau_ci;// MPa
+//d1*(Tmax/(%pi/16)/tau_h)-d1**4=d**4 -- eqn(1)
+Tmax=Tmax*1000;// N.mm
+p=[1 0 0 -Tmax/(%pi*tau_h/16) -d**4] ;// polynomial coefficients from eqn(1)
+d1=roots(p);// roots of poly 
+d1=d1(1);// mm (taking +ve value)
+d1=100;// mm (empirically adopted) 
+t1=(d1-d)/2;// mm (thickness of hub)
+printf('\n thickness of hub = %.f mm',t1)
+d4=d+t1;// mm (diameter of recess in flanges)
+printf('\n diameter of recess in flanges = %.f mm',d4)
+d3=4*d;// mm (outside diameter of protecting flange)
+printf('\n outside diameter of protecting flange = %.f mm',d3)
+
+// Hub length
+b=d/4;// mm (width of key)
+l=1.5*d;// mm (length of key)
+printf('\n width of key = %.1f mm. Use b = 15 mm',b)
+b=15;// mm
+printf('\n length of key = %.f mm.',l) 
+t=b;// mm (thickness for square key)
+printf('\n thickness for square key = %.f mm',t)
+printf('\n Hub length = %.f mm',l)
+
+//Number of bolts
+n=floor(4*d/150+3);// no. of bolts
+printf('\n Number of bolts = %.f',n)
+
+// Bolt diameter
+r2=1.5*d;// mm
+F=Tmax/r2/n;// N
+//%pi/4*db**2*tau_b=F
+db=sqrt(F/(%pi/4*tau_b));// mm
+printf('\n Bolt diameter = %.2f mm. Use db=12 mm',db)
+bolt_dia=db;//mm
+
+// Bolt diameter based on Tensile load
+r3=d3/2;// mm
+r4=d4/2;// mm
+rf=2/3*((r3**3-r4**3)/(r3**2-r4**2));// mm
+//Tmax=n*mu*Pi*rf;// N
+Pi=Tmax/(n*mu*rf);// N
+// Pi=%pi/4*db**2*sigma_t
+sigma_t=sigma_b;// MPa
+db=sqrt(Pi/(%pi/4*sigma_t));// mm 
+printf('\n Bolt diameter (based on Tensile load) = %.1f mm. Use db=15 mm',db)
+db=15;// mm (adopted)
+
+// Flange thickness
+t2=0.5*t1+6;// mm (empirically)
+printf('\n Flange thickness = %.1f mm. Use t=20 mm',t2)
+t2=20;// mm (adopted)
+//F=n*db*t2*sigma_c
+sigma_ci=F/n/db/t2;// MPa
+//2*%pi*d1**2*tau*t2/4=Tmax
+tau=Tmax/(2*%pi*d1**2*t2/4);// MPa
+printf('\n permissible bearing stress in flange = %.2f MPa < 30 MPa',sigma_ci)
+printf('\n shearing of the flange at the junction with hub = %.2f MPa < 15 MPa.',tau)
+printf(' Values are acceptable.')
+
+// Check for crushing of bolt
+//n*db*t2*sigma_cb*d2/2=Tmax
+d2=d1+d;// mm
+db=bolt_dia;//mm
+sigma_cb=Tmax/(n*db*t2*d2/2);// MPa
+printf('\n permissible crushing strength of bolts = %.1f MPa < 60 MPa.',sigma_cb)
+printf(' Hence design is safe.')
+
+// Thickness of protecting flange
+t3=0.5*t2;// mm
+printf('\n Thickness of protecting flange = %.f mm', t3)
+// Hub overlap 
+ho=3;// mm (min)
+printf('\n Hub overlap = %.f mm (min)',ho)
+//Note - Answer for **Bolt diameter based on Tensile load** is calculated wrong in the textbook(error in Pi calculation).