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+// exa 8.12 Pg 236
+clc;clear;close;
+
+// Given Data
+d=6;//mm
+Do=75;// mm
+tau=350;// N/mm.sq.
+G=84;// kN/mm.sq.
+
+printf('\n (i) neglecting the effect of curvature')
+dm=Do-d;// mm
+C=dm/d;// spring index
+Ks=1+0.5/C;// shear stress factor
+//tau=Ks*(8*Fmax*C)/(%pi*d**2)
+Fmax=tau/(Ks*(8*C)/(%pi*d**2));// N
+printf('\n Axial load = %.1f N',Fmax)
+delBYi=8*Fmax*C**3/(G*10**3*d);// mm/turn
+printf('\n deflection per active turn = %.3f mm/turn',delBYi)
+printf('\n\n (ii) considering the effect of curvature')
+Kw=(4*C-1)/(4*C-4)+0.615/C;// Wahl's correction factor
+//tau=Kw*(8*Fmax*C)/(G*d)
+Fmax=tau/(Kw*8*C/(%pi*d**2));
+printf('\n Axial load = %.1f N',Fmax)
+delBYn=8*Fmax*C**3/(G*10**3*d);// mm/turn
+printf('\n deflection per active turn = %.3f mm/turn',delBYn)
+// Note - answer in the textbook is wrong for last part.