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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
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