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Diffstat (limited to '3250/CH3/EX3.9/Ex3_9.sce')
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1 files changed, 26 insertions, 0 deletions
diff --git a/3250/CH3/EX3.9/Ex3_9.sce b/3250/CH3/EX3.9/Ex3_9.sce new file mode 100755 index 000000000..7d5657b38 --- /dev/null +++ b/3250/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,26 @@ +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 +// Sample Problem 9 on page no. 135 +printf("\n # PROBLEM 3.9 # \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 +alpha = 90 /((12*(Rp+0.45*t)*M/(E*(10^3)*(t^3)))+1) +Ls = 2*(((Rp+0.45*t)*%pi/4) + 50-(Rp+t)) +printf("\n Maximum bending force = %d N, \n The required puch angle = %f°,\n The stock length = %f mm",Fmax_,alpha,Ls) +// Answer in the book for maximum bending force is given as 4144 N + |