initial commit / add all books

24 files changed, 401 insertions, 0 deletions

diff --git a/3250/CH3/EX3.1/Ex3_1.sce b/3250/CH3/EX3.1/Ex3_1.sce
new file mode 100755
index 000000000..eadcb5ec9
--- /dev/null
+++ b/3250/CH3/EX3.1/Ex3_1.sce
@@ -0,0 +1,19 @@
+clc 

+// Given that

+A = 150*6 // Cross-section of strips in mm^2

+ti = 6 // Thickness in mm

+pA = 0.20 // Reduction in area

+d = 400 // Diameter of steel rolls in mm

+Ys = 0.35// Shear Yield stress of the material before rolling in KN/mm^2

+Ys_ = 0.4// Shear Yield stress of the material after rolling in KN/mm^2

+mu = 0.1 // Cofficient of friction

+// Sample Problem 1 on page no. 112

+printf("\n # PROBLEM 3.1 # \n")

+tf =0.8*ti

+Ys_a = (Ys + Ys_)/2

+r=d/2

+thetaI = sqrt((ti-tf)/r)

+lambdaI=2*sqrt(r/tf)*atan(thetaI *sqrt(r/tf))

+lambdaN = (1/2)*((1/mu)*(log(tf/ti)) + lambdaI)

+thetaN  =(sqrt(tf/r))*(tan((lambdaN/2)*(sqrt(tf/r))))

+printf("\n The final srip thickness is %f mm, \n The avg shear yield stress during the process is %f KN/mm^2, \n The angle subtended by the deformation zone at the roll centre is %f rad, \n The location of neutral point is %f rad.",tf,Ys_a,thetaI,thetaN)

diff --git a/3250/CH3/EX3.1/Ex3_1.txt b/3250/CH3/EX3.1/Ex3_1.txt
new file mode 100755
index 000000000..ac4331849
--- /dev/null
+++ b/3250/CH3/EX3.1/Ex3_1.txt
@@ -0,0 +1,6 @@
+ # PROBLEM 3.1 # 

+

+ The final srip thickness is 4.800000 mm, 

+ The avg shear yield stress during the process is 0.375000 KN/mm^2, 

+ The angle subtended by the deformation zone at the roll centre is 0.077460 rad, 

+ The location of neutral point is 0.022685 rad. 
\ No newline at end of file
diff --git a/3250/CH3/EX3.10/Ex3_10.sce b/3250/CH3/EX3.10/Ex3_10.sce
new file mode 100755
index 000000000..a66d7c1de
--- /dev/null
+++ b/3250/CH3/EX3.10/Ex3_10.sce
@@ -0,0 +1,30 @@
+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
+
+ 
diff --git a/3250/CH3/EX3.10/Ex3_10.txt b/3250/CH3/EX3.10/Ex3_10.txt
new file mode 100755
index 000000000..0aa9e6ee5
--- /dev/null
+++ b/3250/CH3/EX3.10/Ex3_10.txt
@@ -0,0 +1,5 @@
+ 

+ # PROBLEM 3.10 # 

+

+ Minimum value of die length = 34.367396 mm, 

+ Minimum required capacity of the machine = 2313 N 
\ No newline at end of file
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
+
+ 
diff --git a/3250/CH3/EX3.11/Ex3_11.txt b/3250/CH3/EX3.11/Ex3_11.txt
new file mode 100755
index 000000000..18883bd67
--- /dev/null
+++ b/3250/CH3/EX3.11/Ex3_11.txt
@@ -0,0 +1,4 @@
+ # PROBLEM 3.11 # 

+

+ Maximum force required for extruding the cylindrical aluminium billet = 2436266 N, 

+ Percent of the total power input will be lost in friction at the start of the operation = 66.024761 percent.  
\ No newline at end of file
diff --git a/3250/CH3/EX3.12/Ex3_12.sce b/3250/CH3/EX3.12/Ex3_12.sce
new file mode 100755
index 000000000..9d8acf773
--- /dev/null
+++ b/3250/CH3/EX3.12/Ex3_12.sce
@@ -0,0 +1,15 @@
+clc 
+// Given that
+d = 50 // Diameter of the steel sheet in mm
+t = 3 // Thickness of the steel sheet in mm
+e = 1.75 // True fracture strain
+Y = 2.1e3 // True fracture stress for the material in N/mm^2
+// Sample Problem 12 on page no. 149
+printf("\n # PROBLEM 3.12 # \n")
+C_0 = (t/(1.36*exp(e)))*((2*exp(e))-1)/((2.3*exp(e))-1)
+p = t*(1/2.45)*((1.9*exp(e))-1)/((2.56*exp(e))-1)
+F = Y*C_0*%pi*d
+W = (1/2)*(F)*(p)*(10^-3)
+printf("\n The proper clearance between die and punch = %f mm, \n Maximum punching force = %f N, \n Energy required to punch the hole = %f J",C_0,F/1000,W)
+// Answer in the book given as 45.74 J for energy required to punch the hole
+ 
diff --git a/3250/CH3/EX3.12/Ex3_12.txt b/3250/CH3/EX3.12/Ex3_12.txt
new file mode 100755
index 000000000..43b4179b3
--- /dev/null
+++ b/3250/CH3/EX3.12/Ex3_12.txt
@@ -0,0 +1,5 @@
+ # PROBLEM 3.12 # 

+

+ The proper clearance between die and punch = 0.329240 mm, 

+ Maximum punching force = 108.605364 N, 

+ Energy required to punch the hole = 48.101933 J 
\ No newline at end of file
diff --git a/3250/CH3/EX3.2/Ex3_2.sce b/3250/CH3/EX3.2/Ex3_2.sce
new file mode 100755
index 000000000..7500232e3
--- /dev/null
+++ b/3250/CH3/EX3.2/Ex3_2.sce
@@ -0,0 +1,50 @@
+clc 
+// Given that
+A = 150*6 // Cross-section of strips in mm^2
+w = 150 // Width of the strip in mm
+ti = 6 // Thickness in mm
+pA = 0.20 // Reduction in area
+d = 400 // Diameter of steel rolls in mm
+Ys = 0.35// Shear Yield stress of the material before rolling in KN/mm^2
+Ys_ = 0.4// Shear Yield stress of the material after rolling in KN/mm^2
+mu = 0.1 // Cofficient of friction
+v = 30 // Speed of rolling in m/min
+// Sample Problem 2 on page no. 113
+printf("\n # PROBLEM 3.2 # \n")
+tf =0.8*ti
+Ys_a = (Ys + Ys_)/2
+r=d/2
+thetaI = sqrt((ti-tf)/r)
+lambdaI=2*sqrt(r/tf)*atan(thetaI *sqrt(r/tf))
+lambdaN = (1/2)*((1/mu)*(log(tf/ti)) + lambdaI)
+thetaN  =(sqrt(tf/r))*(tan((lambdaN/2)*(sqrt(tf/r))))
+Dtheta_a = thetaN/4
+Dtheta_b = (thetaI- thetaN)/8
+printf("The values of P_after are\n")
+i = 0
+for i = 0:4
+    theta = i*Dtheta_a
+    y = (1/2)* (tf+r*theta^2)
+    lambda = 2*sqrt(r/tf)*atand(theta*(%pi/180) *sqrt(r/tf))
+    p_a = 2*Ys_a*(2*y/tf)*(exp(mu*lambda))
+    printf("%f \n",p_a)
+end
+I1 = (Dtheta_a/3) *(0.75+.925+4*(.788+.876)+2*.830)// By Simpson's rule
+printf("The values of P_before are\n")
+for i = 0:8
+    theta1 = i*Dtheta_b + thetaN
+    y = (1/2)* (tf+r*theta1^2)
+    lambda = 2*sqrt(r/tf)*atand(theta1*(%pi/180) *sqrt(r/tf))
+    p_b = 2*Ys_a*(2*y/ti)*(exp(mu*(lambdaI-lambda)))
+    printf(" %f \n",p_b)
+end
+I2 = (Dtheta_b/3)*(0.925+.75+4*(.887+.828+.786+.759) + 2*(.855+.804+.772))//By Simpson's rule
+F = r*(I1 + I2)
+F_ = F*w
+T = (r^2)*mu*(I2-I1)
+T_ =T*w
+W = v*(1000/60)/r
+P = 2*T_*W
+printf("\n The roll separating force = %d kN,\n The power required in the rolling process = %f kW",ceil(F_),P/1000)
+// Answer in the book for the power required in the rolling process is given as 75.6 kW
+ 
diff --git a/3250/CH3/EX3.2/Ex3_2.txt b/3250/CH3/EX3.2/Ex3_2.txt
new file mode 100755
index 000000000..dc85fe62b
--- /dev/null
+++ b/3250/CH3/EX3.2/Ex3_2.txt
@@ -0,0 +1,21 @@
+ 

+ # PROBLEM 3.2 # 

+The values of P_after are

+0.750000 

+0.787351 

+0.828770 

+0.874670 

+0.925500 

+The values of P_before are

+ 0.923035 

+ 0.884560 

+ 0.850663 

+ 0.820780 

+ 0.794406 

+ 0.771086 

+ 0.750418 

+ 0.732039 

+ 0.715627 

+

+ The roll separating force = 1908 kN,

+ The power required in the rolling process = 77.376678 kW 
\ No newline at end of file
diff --git a/3250/CH3/EX3.3/Ex3_3.sce b/3250/CH3/EX3.3/Ex3_3.sce
new file mode 100755
index 000000000..fbb8cb052
--- /dev/null
+++ b/3250/CH3/EX3.3/Ex3_3.sce
@@ -0,0 +1,50 @@
+clc 
+// Given that
+A = 150*6 // Cross-section of strips in mm^2
+w = 150 // Width of the strip in mm
+ti = 6 // Thickness in mm
+pA = 0.20 // Reduction in area
+d = 400 // Diameter of steel rolls in mm
+Ys = 0.35// Shear Yield stress of the material before rolling in KN/mm^2
+Ys_ = 0.4// Shear Yield stress of the material after rolling in KN/mm^2
+mu = 0.1 // Cofficient of friction
+mu_ = 0.005 // Cofficient of friction in bearing 
+D = 150 // The diameter of bearing in mm
+v = 30 // Speed of rolling in m/min
+// Sample Problem 3 on page no. 115
+printf("\n # PROBLEM 3.3 # \n")
+tf =0.8*ti
+Ys_a = (Ys + Ys_)/2
+r=d/2
+thetaI = sqrt((ti-tf)/r)
+lambdaI=2*sqrt(r/tf)*atan(thetaI *sqrt(r/tf))
+lambdaN = (1/2)*((1/mu)*(log(tf/ti)) + lambdaI)
+thetaN  =(sqrt(tf/r))*(tan((lambdaN/2)*(sqrt(tf/r))))
+Dtheta_a = thetaN/4
+Dtheta_b = (thetaI- thetaN)/8
+i = 0
+for i = 0:4
+    theta = i*Dtheta_a
+    y = (1/2)* (tf+r*theta^2)
+    lambda = 2*sqrt(r/tf)*atand(theta*(%pi/180) *sqrt(r/tf))
+    p_a = 2*Ys_a*(2*y/tf)*(exp(mu*lambda))
+end
+I1 = (Dtheta_a/3) *(0.75+.925+4*(.788+.876)+2*.830)
+for i = 0:8
+    theta1 = i*Dtheta_b + thetaN
+    y = (1/2)* (tf+r*theta1^2)
+    lambda = 2*sqrt(r/tf)*atand(theta1*(%pi/180) *sqrt(r/tf))
+    p_b = 2*Ys_a*(2*y/ti)*(exp(mu*(lambdaI-lambda)))
+end
+I2 = (Dtheta_b/3)*(0.925+.75+4*(.887+.828+.786+.759) + 2*(.855+.804+.772))
+F = r*(I1 + I2)
+F_ = F*w
+T = (r^2)*mu*(I2-I1)
+T_ =T*w
+W = v*(1000/60)/r
+P_ = 2*T_*W
+Pl = mu_*F_*D*W
+P = Pl+P_
+printf("\n The mill power = %f kW",P/1000)
+// Answer in the book is given as 79.18 kW
+ 
diff --git a/3250/CH3/EX3.3/Ex3_3.txt b/3250/CH3/EX3.3/Ex3_3.txt
new file mode 100755
index 000000000..48ecc32cc
--- /dev/null
+++ b/3250/CH3/EX3.3/Ex3_3.txt
@@ -0,0 +1,3 @@
+ # PROBLEM 3.3 # 

+

+ The mill power = 80.952339 kW 
\ No newline at end of file
diff --git a/3250/CH3/EX3.4/Ex3_4.sce b/3250/CH3/EX3.4/Ex3_4.sce
new file mode 100755
index 000000000..d41dfae48
--- /dev/null
+++ b/3250/CH3/EX3.4/Ex3_4.sce
@@ -0,0 +1,28 @@
+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 
+
+ 
diff --git a/3250/CH3/EX3.4/Ex3_4.txt b/3250/CH3/EX3.4/Ex3_4.txt
new file mode 100755
index 000000000..860a80a1d
--- /dev/null
+++ b/3250/CH3/EX3.4/Ex3_4.txt
@@ -0,0 +1,3 @@
+ # PROBLEM 3.4 # 

+

+ The maximum forging force = 4.890305e+05 N 
\ No newline at end of file
diff --git a/3250/CH3/EX3.5/Ex3_5.sce b/3250/CH3/EX3.5/Ex3_5.sce
new file mode 100755
index 000000000..9bce6c835
--- /dev/null
+++ b/3250/CH3/EX3.5/Ex3_5.sce
@@ -0,0 +1,23 @@
+clc 
+// Given that
+mu = 0.08// 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 5 on page no. 119
+printf("\n # PROBLEM 3.5 # \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
+I = intg(0,l,f)
+F = 2*(I)
+F_ = F*L
+printf("\n The maximum forging force = %e N",F_)
+
+ 
diff --git a/3250/CH3/EX3.5/Ex3_5.txt b/3250/CH3/EX3.5/Ex3_5.txt
new file mode 100755
index 000000000..2c4178dc6
--- /dev/null
+++ b/3250/CH3/EX3.5/Ex3_5.txt
@@ -0,0 +1,4 @@
+ 

+ # PROBLEM 3.5 # 

+

+ The maximum forging force = 2.361194e+05 N 
\ No newline at end of file
diff --git a/3250/CH3/EX3.6/Ex3_6.sce b/3250/CH3/EX3.6/Ex3_6.sce
new file mode 100755
index 000000000..7469536ab
--- /dev/null
+++ b/3250/CH3/EX3.6/Ex3_6.sce
@@ -0,0 +1,23 @@
+clc 
+// Given that
+r = 150 // Radius of the circular disc of lead in mm
+Ti = 50 // Initial thickness of the disc in mm
+Tf = 25 // Reduced thickness of the disc in mm
+mu = 0.25// Cofficient of friction between the job and the dies 
+K = 4 // Avg shear yield stress of the lead in N/mm^2
+// Sample Problem 6 on page no. 122
+printf("\n # PROBLEM 3.6 # \n")
+R = r*sqrt(2)
+rs = (R - ((Tf/(2*mu)) * log(1/(mu*sqrt(3)))))
+funcprot(0)
+function p1 = f(x), p1 = (((sqrt(3))*K)*exp((2*mu/Tf)*(R-x)))*x,
+endfunction
+I = intg(rs,R,f)
+funcprot(0)
+function p2 = f(y), p2 = ((2*K/Tf)*(R-y) + ((K/mu)*(1+log(mu*sqrt(3)))))*y,
+endfunction
+I_ = intg(0,rs,f)
+F = 2*%pi*(I+I_)
+printf("\n The maximum forging force = %e N",F)
+
+ 
diff --git a/3250/CH3/EX3.6/Ex3_6.txt b/3250/CH3/EX3.6/Ex3_6.txt
new file mode 100755
index 000000000..9ddcad016
--- /dev/null
+++ b/3250/CH3/EX3.6/Ex3_6.txt
@@ -0,0 +1,4 @@
+

+ # PROBLEM 3.6 # 

+

+ The maximum forging force = 3.648809e+06 N 
\ No newline at end of file
diff --git a/3250/CH3/EX3.7/Ex3_7.sce b/3250/CH3/EX3.7/Ex3_7.sce
new file mode 100755
index 000000000..82270e06f
--- /dev/null
+++ b/3250/CH3/EX3.7/Ex3_7.sce
@@ -0,0 +1,23 @@
+clc 
+// Given that
+Di = 12.7 // Intial diameter in mm
+Df = 10.2 // Final diameter in mm
+v = 90 // Drawn speed in m/min
+alpha=6 // Half angle of dia in degree
+mu = 0.1// Cofficient of friction between the job and the dies 
+Y = 207 // Tensile  yield stress of the steel specimen in N/mm^2
+Y_ = 414 // Tensile yield stress of the similar specimen at strain 0.5 in N/mm^2
+e = 0.5 // Strain
+// Sample Problem 7 on page no. 126
+printf("\n # PROBLEM 3.7 # \n")
+e_ =2* log(Di/Df)
+Y_e = Y + (Y_ - Y)*e_/e
+Y__ = (Y+Y_e)/2
+phi = 1 + (mu/tand(alpha))
+Y_f = Y__ * ((phi/(phi-1)) *(1-((Df/Di)^(2*(phi-1)))))
+p = Y_f * (%pi/4)*(Df^2)*v/60
+Dmax = 1- (1/(phi^(1/(phi-1))))
+printf("\n Drawing power = %f kW, \n The maximum passible reduction with same die = %f mm",p/1000,Dmax)
+
+
+ 
diff --git a/3250/CH3/EX3.7/Ex3_7.txt b/3250/CH3/EX3.7/Ex3_7.txt
new file mode 100755
index 000000000..da97d764e
--- /dev/null
+++ b/3250/CH3/EX3.7/Ex3_7.txt
@@ -0,0 +1,5 @@
+ 

+ # PROBLEM 3.7 # 

+

+ Drawing power = 25.530385 kW, 

+ The maximum passible reduction with same die = 0.504749 mm 
\ No newline at end of file
diff --git a/3250/CH3/EX3.8/Ex3_8.sce b/3250/CH3/EX3.8/Ex3_8.sce
new file mode 100755
index 000000000..16904a775
--- /dev/null
+++ b/3250/CH3/EX3.8/Ex3_8.sce
@@ -0,0 +1,20 @@
+clc 
+// Given that
+Ri = 30 // Inside radius of cup in mm
+t = 3 // Thickness in mm
+Rb = 40 // Radius of the blank in mm
+K = 210 // Shear yield stress of the material in N/mm^2
+Y = 600 // Maximum allowable stress in N/mm^2
+Beta = 0.05
+mu = 0.1// Cofficient of friction between the job and the dies 
+// Sample Problem 8 on page no. 130
+printf("\n # PROBLEM 3.8 # \n")
+Fh = Beta*%pi*(Rb^2)*K
+Y_r = (mu*Fh/(%pi*Rb*t))+(2*K*log(Rb/Ri))
+Y_z = Y_r*exp(mu*%pi/2)
+F = 2*%pi*Ri*t*Y_z
+Y_r_ = Y/exp(mu*%pi/2)
+Rp = (Rb/exp((Y_r_/(2*K)) - ((mu*Fh)/(2*%pi*K*Rb*t))))-t
+printf("\n Drawing force = %d N, \n Minimum passible radius of the cup which can drawn from the given blank without causing a fracture = %f mm",F,Rp)
+// Answer in the book given as 62680 N
+
diff --git a/3250/CH3/EX3.8/Ex3_8.txt b/3250/CH3/EX3.8/Ex3_8.txt
new file mode 100755
index 000000000..18bb5f855
--- /dev/null
+++ b/3250/CH3/EX3.8/Ex3_8.txt
@@ -0,0 +1,4 @@
+ # PROBLEM 3.8 # 

+

+ Drawing force = 89210 N, 

+ Minimum passible radius of the cup which can drawn from the given blank without causing a fracture = 9.198393 mm 
\ No newline at end of file
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
+ 
diff --git a/3250/CH3/EX3.9/Ex3_9.txt b/3250/CH3/EX3.9/Ex3_9.txt
new file mode 100755
index 000000000..4e5f2a73e
--- /dev/null
+++ b/3250/CH3/EX3.9/Ex3_9.txt
@@ -0,0 +1,5 @@
+ # PROBLEM 3.9 # 

+

+ Maximum bending force = 4124 N, 

+ The required puch angle = 88.681608°,

+ The stock length = 89.175451 mm 
\ No newline at end of file