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+clc 
+// Given that
+L_ = 20 // Length of the mild steel product in mm
+h = 50 // Height of the mild steel product in mm
+L = 50 // Horizontal length of the mild steel product in mm
+t = 5 // Thickness in mm
+l=25 // Length of the bend in mm
+E = 207 // Modulus of elasticity in kN/mm^2
+n = 517 // Strain hardening rate in N/mm^2
+Y = 345 // Yield stress in N/mm^2
+mu = 0.1// Cofficient of friction
+e = 0.2 // Fracture strain
+theta = 20 // Bend angle in degree
+F = 3000 // Maximum available force in N
+// Sample Problem 10 on page no. 136
+printf("\n # PROBLEM 3.10 # \n")
+Rp = ((1 /((exp(e) - 1)))-0.82)*t/1.82
+Y_1 = Y+n*e
+Y_2 = Y + n*(log(1+(1/(2.22*(Rp/t)+1))))
+M = ((0.55*t)^2)*((Y/6)+(Y_1/3)) + ((0.45*t)^2)*((Y/6)+(Y_2/3))
+Fmax = (M/l)*(1+(cosd((atand(mu))+mu*sind(atand(mu)))))
+Fmax_ = L_*Fmax
+lmin  = Fmax_*l/F
+Ls = 2*(((Rp+0.45*t)*%pi/4) + 50-(Rp+t))
+lmax = Ls / 2
+Fmax_min  = Fmax_*l/lmax
+printf("\n Minimum value of die length = %f mm, \n Minimum required capacity of the machine = %d N",lmin,ceil(Fmax_min))
+// Answer in the book is give as 2323 N for Minimum required capacity of the machine
+
+