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+clc
+// Given that
+mu = 0.25 // Cofficient of friction between the job and the dies
+Y = 7 // Avg yield stress of the lead in N/mm^2
+h = 6 // Height of die in mm
+L = 150 // Length of the strip in mm
+V1 = 24*24*150 // Volume of the strip in mm^3
+V2 = 6*96*150 // Volume of the die in mm^3
+w= 96 // Weidth of the die in mm
+// Sample Problem 4 on page no. 118
+printf("\n # PROBLEM 3.4 # \n")
+K = Y/sqrt(3)
+x_ = (h/(2*mu))*(log(1/(2*mu)))
+l = w/2
+funcprot(0)
+function p1 = f(x), p1 = (2*K)*exp((2*mu/h)*x),
+endfunction
+funcprot(0)
+I1 = intg(0,x_,f)
+function p2 = f(y), p2=(2*K)*((1/2*mu)*(log(1/(2*mu))) + (y/h)),
+endfunction
+I2 = intg(x_,l,f)
+F = 2*(I1+I2)
+F_ = F*L
+printf("\n The maximum forging force = %e N",F_)
+// Answer in the book is given as 0.54*10^6 N
+
+