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Diffstat (limited to '3250/CH3/EX3.11/Ex3_11.sce')
| -rwxr-xr-x | 3250/CH3/EX3.11/Ex3_11.sce | 25 |
1 files changed, 25 insertions, 0 deletions
diff --git a/3250/CH3/EX3.11/Ex3_11.sce b/3250/CH3/EX3.11/Ex3_11.sce new file mode 100755 index 000000000..9b18113b1 --- /dev/null +++ b/3250/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,25 @@ +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 + + |