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+// exa 8.8 Pg 234
+clc;clear;close;
+
+// Given Data
+Fmin=0;// N
+Fmax=1000;// N
+del=80;// mm
+Do=25;// mm
+n=30;// no. of turns
+G=85;// kN/mm.sq.
+
+k=(Fmax-Fmin)/del;// N/mm (Spring stiffness)
+printf('\n Spring stiffness = %.1f N/mm',k)
+// k=G*d/(8*C**3*n) (Spring stiffness)
+C_cube_BY_d=G*10**3/(k*8*n);// 
+
+function [C,d]=hitntrial(c3d,Do)
+    for C=5:-0.1:4.5
+        d=C**3/(c3d);
+        Doo=d*C+C;
+        if Doo<Do 
+            break;
+        end;
+    end
+
+endfunction;
+
+[C,d]=hitntrial(C_cube_BY_d,Do)
+printf('\n By hit and trial method and using value of C^3/d -')
+printf('\n value of Spring Index, C = %.1f ',C)
+printf('\n value of wire diameter, d = %.1f mm',d)
+printf('\n But we adopt d=4 mm.')
+d=4;// mm (adopted for design)
+C=(C_cube_BY_d*d)**(1/3);// Spring index
+printf('Hence, Spring Index = %.2f ',C)
+Dm=C*d;// mm
+printf('\n Mean coil diameter = %.2f mm', Dm)
+Do=Dm+d;// mm
+printf('\n Outside coil diameter = %.2f mm < 25 mm. Hence design is ok.', Do)
+Kw=(4*C-1)/(4*C-4)+0.615/C;// Wahl's correction factor
+printf('\n Wahl''s correction factor = %.3f ',Kw)
+tau=8*Kw*C*Fmax/(%pi*d**2);// N/mm.sq.
+printf('\n Maximum shear stress = %.2f N/mm.sq.',tau)