initial commit / add all books

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diff --git a/3250/CH3/EX3.11/Ex3_11.sce b/3250/CH3/EX3.11/Ex3_11.sce
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+clc 
+// Given that
+d = 50 // Diameter of the billet in mm
+L =75 // Length of the billet in mm
+D = 10 // Final diameter of billet in mm
+Y = 170 // Avg tensile yield stress for aluminium in N/mm^2
+mu = 0.15 // Cofficient of the friction
+// Sample Problem 11 on page no. 141
+printf("\n # PROBLEM 3.11 # \n")
+l = L - ((d-D)/2)*cotd(45)
+phi = 1+mu
+Y_x = Y*(phi/(phi-1))*(((d/D)^(2*(phi-1)))-1)
+F = (%pi/4)*(d^2)*Y_x + (%pi/sqrt(3))*(d*l*Y)
+Pf = %pi*Y*(d^2)*((phi/(2*mu))*(((d/D)^(2*mu))-1)-log(d/D)) + (%pi/sqrt(3))*Y*d*l
+Loss_f = (Pf/F)*100
+Y_X = Y*4.31*log(d/D)
+F_ = (%pi/4)*(d^2)*Y_X + (%pi/sqrt(3))*(d*l*Y)
+Pf_1 = (%pi/sqrt(3))*Y*(d^2)*(log(d/D))
+Pf_2 = (%pi/sqrt(3))*(d*l*Y)
+Pf_ = Pf_1+Pf_2
+Loss_f_ = (Pf_/F_)*100
+printf("\n Maximum force required for extruding the cylindrical aluminium billet = %d N, \n Percent of the total power input will be lost in friction at the start of the operation = %f percent. ",F,Loss_f_)
+// Answer in the book given as 2436444 N for max force required for extruding the cylindrical aluminium billet
+
+