diff options
Diffstat (limited to '3250')
142 files changed, 1731 insertions, 0 deletions
diff --git a/3250/CH2/EX1.1/Ex2_1.sce b/3250/CH2/EX1.1/Ex2_1.sce new file mode 100755 index 000000000..10d3cdb8c --- /dev/null +++ b/3250/CH2/EX1.1/Ex2_1.sce @@ -0,0 +1,16 @@ +clc +// Given that +h=15 // Height of spur in cm +l= 50 // Length of cast in cm +w= 25 // weidth of cast in cm +h1= 15 // Height of cast in cm +g= 981 // Acceleration due to gravity in cm/sec^2 +Ag= 5 // Cross sectional area of the grate in cm^2 +// Sample Problem 1 on page no. 46 +printf("\n # PROBLEM 2.1 # \n") +v3= sqrt(2* g * h) +V = l*w*h1 +tf1= V/(Ag*v3) +Am = l*w +tf2 = (Am/Ag)*(1/sqrt(2*g))*2*(sqrt(h) - sqrt(h-h1)) +printf("\n Filling time for first design = %f sec, \n Filling time for second design = %f sec", tf1, tf2) diff --git a/3250/CH2/EX1.1/Ex2_1.txt b/3250/CH2/EX1.1/Ex2_1.txt new file mode 100755 index 000000000..44e364a5d --- /dev/null +++ b/3250/CH2/EX1.1/Ex2_1.txt @@ -0,0 +1,4 @@ +# PROBLEM 2.1 #
+
+ Filling time for first design = 21.859294 sec,
+ Filling time for second design = 43.718589 sec
\ No newline at end of file diff --git a/3250/CH2/EX2.2/Ex2_2.sce b/3250/CH2/EX2.2/Ex2_2.sce new file mode 100755 index 000000000..4b9a127c6 --- /dev/null +++ b/3250/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,28 @@ +clc +// Given that +h=15 // Height of spur in cm +l= 50 // Length of cast in cm +w= 25 // weidth of cast in cm +h1= 15 // Height of cast in cm +g= 981 // Acceleration due to gravity in cm/sec^2 +Ag= 5 // Cross sectional area of the grate in cm^2 +Dm = 7800 // Density of molten Fe in Kg/m^3 +Neta = 0.00496 // Kinetic viscosity in Kg/m-sec +theta = 90 // Angle in degree +Eq = 25 // (L/D) Equivalent +// Sample Problem 2 on page no. 53 +printf("\n # PROBLEM 2.2 # \n") +v3= sqrt(2* g * h)*(10^(-2)) +d= sqrt((Ag*4)/(%pi))*(10^(-2)) +Re = Dm*v3*d/Neta +f = 0.0791*(Re)^(-1/4) +L=0.12 // in meter +Cd= (1+0.45+4*f*((L/d)+Eq))^(-1/2) +v3_ = Cd*v3 +Re_ = (v3_/v3)*(Re) +f_ = 0.0791 *(Re_)^(-1/4) +Cd_ = (1+0.46+4*f_*(L/d + Eq))^(-1/2) +v3__ = Cd_*v3 +V = l*w*h1 +tf= (V/(Ag*v3__))*(10^-2) +printf("\n Filling time for first design = %f sec. ", tf) diff --git a/3250/CH2/EX2.2/Ex2_2.txt b/3250/CH2/EX2.2/Ex2_2.txt new file mode 100755 index 000000000..d6ed53e9d --- /dev/null +++ b/3250/CH2/EX2.2/Ex2_2.txt @@ -0,0 +1,4 @@ +
+ # PROBLEM 2.2 #
+
+ Filling time for first design = 31.653919 sec.
\ No newline at end of file diff --git a/3250/CH2/EX2.3/Ex2_3.sce b/3250/CH2/EX2.3/Ex2_3.sce new file mode 100755 index 000000000..fd7509e03 --- /dev/null +++ b/3250/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,26 @@ +clc +// Given that +Hi=1.2 // Initial height in m +H= 0.05 // Height in m +g= 9.81 // Acceleration due to gravity in m/sec^2 +Dm = 2700 // Density of molten metal in Kg/m^3 +Neta = 0.00273 // Kinetic viscosity in Kg/m-sec +d= 0.075 // Diameter in m +D = 1 // Internal diameter of ladle in m +// Sample Problem on page no. 56 +printf("\n # PROBLEM 2.3 # \n") +v3= sqrt(2* g * Hi) +Re = Dm*v3*d/Neta +ef=0.075 +Cd= (1+ef)^(-1/2) +ef_=0.82 +Re_ = (2+ef_)^(-1/2) +v3_ = sqrt(2*g*H) +Re_ = Dm*v3_*d/Neta +At = (%pi/4)*D^2 +An = (%pi/4)*d^2 +Cd= 0.96 +tf= (sqrt(2/g))*(At/An)*(1/Cd)*sqrt(Hi) +m = Dm*An*Cd*sqrt(2*g*Hi) +m_ = Dm*An*Cd*sqrt(2*g*Hi*0.25) +printf("\n Time required to empty the ladle = %f sec, \n Discharge rate are - \n Initially = %f Kg/sec \n When the ladle is 75 percent empty = %f Kg/sec. ",tf,m,m_) diff --git a/3250/CH2/EX2.3/Ex2_3.txt b/3250/CH2/EX2.3/Ex2_3.txt new file mode 100755 index 000000000..38d8a3065 --- /dev/null +++ b/3250/CH2/EX2.3/Ex2_3.txt @@ -0,0 +1,7 @@ +
+ # PROBLEM 2.3 #
+
+ Time required to empty the ladle = 91.596179 sec,
+ Discharge rate are -
+ Initially = 55.563236 Kg/sec
+ When the ladle is 75 percent empty = 27.781618 Kg/sec.
diff --git a/3250/CH2/EX2.5/Ex2_5.sce b/3250/CH2/EX2.5/Ex2_5.sce new file mode 100755 index 000000000..7081419b9 --- /dev/null +++ b/3250/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,28 @@ +clc +// Given that +thetaF= 1540 // Temperature of mould face in degree centigrade +Theta0 = 28 // Initial temperature of mould in Degree centigrade +L= 272e3 // Latent heat of liquid metal in J/Kg +Dm = 7850 // Density of liquid metal in Kg/m^3 +c = 1.17e+3 //Specific heat of sand in J/Kg-K +k = 0.8655 // Conductivity of sand in W/m-K +D= 1600 // Density of sand in Kg/m^3 +h = 0.1 // Height in m +b = 10 // Thickness of slab in cm +r =h/2// V/A in meter +// Sample Problem 5 on page no. 66 +printf("\n # PROBLEM 2.5 # \n") +lambda = (thetaF - Theta0)*(D*c)/(Dm*L) +Beta1 = 2*lambda/sqrt(%pi) +Alpha = k /(D*c) +ts1 = r^2 /((Beta1^2)*Alpha)//In sec +ts1_=ts1/3600 // In hour +Beta= poly(0,"Beta"); +p=Beta^2 - lambda*(2/sqrt(%pi))*Beta -lambda/3 +Beta2 = roots(p) +printf(" The value of Beta2 is %f, ",Beta2) +printf("\n We only take the positive value of Beta2 , \n Hence Beta2=1.75") +r1 = r/3 +ts2 = (r1^2)/((1.75^2)*Alpha) // in sec +ts2_=ts2/3600//in Hour +printf("\n\n Solidification time for slab-shaped casting = %f hr,\n Solidification time for sphere = %f hr", ts1_,ts2_) diff --git a/3250/CH2/EX2.5/Ex2_5.txt b/3250/CH2/EX2.5/Ex2_5.txt new file mode 100755 index 000000000..fa46acfd9 --- /dev/null +++ b/3250/CH2/EX2.5/Ex2_5.txt @@ -0,0 +1,8 @@ +
+ # PROBLEM 2.5 #
+ The value of Beta2 is 1.748516, The value of Beta2 is -0.252713,
+ We only take the positive value of Beta2 ,
+ Hence Beta2=1.75
+
+ Solidification time for slab-shaped casting = 0.671318 hr,
+ Solidification time for sphere = 0.054495 hr
\ No newline at end of file diff --git a/3250/CH2/EX2.6/Ex2_6.sce b/3250/CH2/EX2.6/Ex2_6.sce new file mode 100755 index 000000000..16eafd4d4 --- /dev/null +++ b/3250/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,24 @@ +clc +// Given that +thetaF= 1540 // Temperature of mould face in degree centigrate +ThetaO = 28 // Initial temperature of mould in Degree centigrate +L= 272e3 // Latent heat of iron in J/Kg +Dm = 7850 // Density of iron in Kg/m^3 +Cs = 0.67e+3 //Specific heat of iron in J/Kg-K +C = 0.376e3 //Specific heat of copper in J/Kg-K +Ks = 83 // Conductivity of iron in W/m-K +K = 398 // Conductivity of copper in W/m-K +D= 8960 // Density of copper in Kg/m^3 +h = .1 // Height in m +// Sample Problem 6 on page no. 73 +printf("\n # PROBLEM 2.6 # \n") +zeta1=0.98//By solving eqauation- zeta*exp(zeta^2)*erf(zeta)=((thetaF-thetaO)*Cs)/(sqrt(pi)*L), zeta = 0.98 +AlphaS = Ks /(Dm*Cs) +ts1 = h^2 / (16*(zeta1^2) * AlphaS)//In sec +ts1_=ts1/3600 // In hour +Phi = sqrt((Ks*Dm*Cs)/(K*D*C)) +zeta2=0.815//By solving eqauation- zeta*exp(zeta^2)*(erf(zeta)+Phi)=((thetaF-thetaO)*Cs)/(sqrt(pi)*L), zeta = 0.815 +ts2 = h^2 / (16*(zeta2^2) * AlphaS)//In sec +ts2_=ts2/3600 // In hour +thetaS= (thetaF-(L*(sqrt(%pi))*zeta2*(exp(zeta2^2))*erf(zeta2))/Cs) +printf("\n Solidification time for slab-shaped casting when the casting is done in a water cooled copper mould = %f hr,\n Solidification time for slab-shaped casting when the casting is done in a very thick copper mould = %f hr,\n The surface temperature of the mould = %f° C", ts1_,ts2_,thetaS) diff --git a/3250/CH2/EX2.6/Ex2_6.txt b/3250/CH2/EX2.6/Ex2_6.txt new file mode 100755 index 000000000..8a8128a8d --- /dev/null +++ b/3250/CH2/EX2.6/Ex2_6.txt @@ -0,0 +1,6 @@ +
+ # PROBLEM 2.6 #
+
+ Solidification time for slab-shaped casting when the casting is done in a water cooled copper mould = 0.011455 hr,
+ Solidification time for slab-shaped casting when the casting is done in a very thick copper mould = 0.016563 hr,
+ The surface temperature of the mould = 684.364285° C
\ No newline at end of file diff --git a/3250/CH2/EX2.7/Ex2_7.sce b/3250/CH2/EX2.7/Ex2_7.sce new file mode 100755 index 000000000..6262bde83 --- /dev/null +++ b/3250/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,27 @@ +clc +// Given that +thetaF= 1540 // Temperature of mould face in degree centigrade +thetaO = 28 // Initial temperature of mould in Degree centigrade +L= 272e3 // Latent heat of iron in J/Kg +Dm = 7850 // Density of iron in Kg/m^3 +Cs = 0.67e+3 //Specific heat of iron in J/Kg-K +C = 0.376e3 //Specific heat of copper in J/Kg-K +Ks = 83 // Conductivity of iron in W/m-K +K = 398 // Conductivity of copper in W/m-K +D= 8960 // Density of copper in Kg/m^3 +h = .1 // Height in m +hF = 1420 // Total heat transfer coefficient across the casting-mould interface in W/m^2-°C +// Sample Problem 7 on page no. 75 +printf("\n # PROBLEM 2.7 # \n") +AlphaS = K /(D*C) +thetaS = 982 //In °C as in example 2.6 +h1= (1+(sqrt((Ks*Dm*Cs)/(K*D*C))))*hF +a = 1/2 + (sqrt((1/4)+Cs*(thetaF-thetaS)/(3*L))) +delta=h/2 +ts = (delta+((h1*delta^2)/(2*Ks)))/((h1*(thetaF-thetaS))/(Dm*L*a)) // in sec +ts_ = ts/3600 // in hours +h2= (1+(sqrt((K*D*C)/(Ks*Dm*Cs))))*hF +gama= ((h2^2)/(K^2))*AlphaS*ts +thetaS_ = thetaO + (thetaS-thetaO)*(1-((exp(gama))*(1-(erf(sqrt(gama)))))) +printf("\n Solidification time = %f hr,\n The surface temperature of the mould = %f ° C", ts_,thetaS_) +// The value of the surface temperature of the mould in the book is given as 658.1° C, Which is wrong. diff --git a/3250/CH2/EX2.7/Ex2_7.txt b/3250/CH2/EX2.7/Ex2_7.txt new file mode 100755 index 000000000..a4194a318 --- /dev/null +++ b/3250/CH2/EX2.7/Ex2_7.txt @@ -0,0 +1,4 @@ + # PROBLEM 2.7 #
+
+ Solidification time = 0.053443 hr,
+ The surface temperature of the mould = 672.151154 ° C
\ No newline at end of file diff --git a/3250/CH2/EX2.8/Ex2_8.sce b/3250/CH2/EX2.8/Ex2_8.sce new file mode 100755 index 000000000..1f7d45ce6 --- /dev/null +++ b/3250/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,28 @@ +clc +// Given that +A= 60*7.5 // Cross sectional area in cm^2 +v=0.05 // Withdrawal rate in m/sec +t = 0.0125 // Thickness in m +thetaF= 1500 // Temperature of mould face in degree centigrate +thetaP = 1550 // +thetaO = 20 // Initial temperature of mould in Degree centigrate +L= 268e3 // Latent heat of molten metal in J/Kg +Dm = 7680 // Density of molten metal in Kg/m^3 +Cs = 0.67e+3 //Specific heat of molten metal in J/Kg-K +Cm = 0.755e3 //Specific heat of mould in J/Kg-K +Ks = 76 // Conductivity of molten metal in W/m-K +hF = 1420 // Heat transfer coefficient at the casting-mould interface in W/m^2-°C +Dtheta = 10 // Maximum temperature of cooling water in ° C +// Sample Problem 8 on page no. 77 +printf("\n # PROBLEM 2.8 # \n") +L_ = L+Cm*(thetaP-thetaF) +x=L_ / (Cs*(thetaF-thetaO)) +y= hF*t/Ks +printf(" L_/(Cs(thetaF-thetaO))=%f,\n hF*t/Ks=%f",x,y) +z=0.11 // Where z=hF^2 * lm / (v*Ks*Dm*Cs) +lm= (z*v*Ks*Dm*Cs)/(hF^2) +Z=0.28 // Where Z=Q/(lm*(thetaF-thetaO)*sqrt(lm*v*Dm*Cs*Ks)) +Q = Z*lm*(thetaF-thetaO)*sqrt(lm*v*Dm*Cs*Ks) +m = Q / (4.2e3*Dtheta) +printf("\n The mould length = %f meter,\n The cooling water requirement = %f Kg/sec", lm,m) +// Answer for The cooling water requirement in the book is given as 5.05 Kg/sec, Which is wrong. diff --git a/3250/CH2/EX2.8/Ex2_8.txt b/3250/CH2/EX2.8/Ex2_8.txt new file mode 100755 index 000000000..5e1c496ff --- /dev/null +++ b/3250/CH2/EX2.8/Ex2_8.txt @@ -0,0 +1,6 @@ +
+ # PROBLEM 2.8 #
+ L_/(Cs(thetaF-thetaO))=0.308340,
+ hF*t/Ks=0.233553
+ The mould length = 1.066684 meter,
+ The cooling water requirement = 48.065525 Kg/sec
\ No newline at end of file diff --git a/3250/CH2/EX2.9/Ex2_9.sce b/3250/CH2/EX2.9/Ex2_9.sce new file mode 100755 index 000000000..e9639576d --- /dev/null +++ b/3250/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,21 @@ +clc +// Given that +a = 15 // Side of the aluminium cube in cm +Sh = 0.065 // Volume shrinkage of aluminium during solidification +// Sample Problem 9 on page no. 81 +printf("\n # PROBLEM 2.9 # \n") +Vc = a^3 +Vr = 3*Sh*Vc +h = ((4*Vr)/%pi)^(1/3) +Rr = 6/h // Where Rr= (A/V)r +Rc = 6/a // Where Rc = (A/V)c +printf("(A/V)r=%f, (A/V)c=%f\n Hence Rr is greater than Rc",Rr,Rc) +dmin = 6/Rc +Vr_ = (%pi/4)*dmin^3 +printf("\n With minimum value of d Vr=%d cm^3 .\n This valume is much more than the minimum Vr necessary. \nLet us now consider the top riser when the optimum cylindrical shape is obtained with h=d/2 \nand again (A/V)r = 6/d. However, with a large top riser,\n the cube loses its top surface for the purpose of heat dissipation.",Vr_) +Rc_ = 5/a +dmin_=6/Rc_ +printf("\n d should be greater than or equal to %d cm",dmin_) +Vr__ = (%pi/4)*dmin_^2 *floor(h) +printf("\n The riser volume with minimum diameter is %d cm^3",Vr__) + diff --git a/3250/CH2/EX2.9/Ex2_9.txt b/3250/CH2/EX2.9/Ex2_9.txt new file mode 100755 index 000000000..5262e1e8b --- /dev/null +++ b/3250/CH2/EX2.9/Ex2_9.txt @@ -0,0 +1,11 @@ +
+ # PROBLEM 2.9 #
+(A/V)r=0.636422, (A/V)c=0.400000
+ Hence Rr is greater than Rc
+ With minimum value of d Vr=2650 cm^3 .
+ This valume is much more than the minimum Vr necessary.
+Let us now consider the top riser when the optimum cylindrical shape is obtained with h=d/2
+and again (A/V)r = 6/d. However, with a large top riser,
+ the cube loses its top surface for the purpose of heat dissipation.
+ d should be greater than or equal to 18 cm
+ The riser volume with minimum diameter is 2290 cm^3
\ No newline at end of file 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 diff --git a/3250/CH4/EX4.1/Ex4_1.sce b/3250/CH4/EX4.1/Ex4_1.sce new file mode 100755 index 000000000..0b768c5e9 --- /dev/null +++ b/3250/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,11 @@ +clc +// Given that +alpha = 10 // Rake angle in Degree +t = 0.4 // Chip thickness in mm +T = 0.15 // Uncut chip thickness in mm +// Sample Problem 1 on page no. 187 +printf("\n # PROBLEM 4.1 # \n") +r = T/t +phi = atand((r*cosd(alpha))/(1-r*sind(alpha))) +gama = cotd(phi) + tand(phi-alpha) +printf("\n Shear plane angle = %f°, \n Magnitude of the shear strain = %f",phi,gama) diff --git a/3250/CH4/EX4.1/Ex4_1.txt b/3250/CH4/EX4.1/Ex4_1.txt new file mode 100755 index 000000000..cb6cb9d65 --- /dev/null +++ b/3250/CH4/EX4.1/Ex4_1.txt @@ -0,0 +1,4 @@ +# PROBLEM 4.1 #
+
+ Shear plane angle = 21.555321°,
+ Magnitude of the shear strain = 2.735935
\ No newline at end of file diff --git a/3250/CH4/EX4.10/Ex4_10.sce b/3250/CH4/EX4.10/Ex4_10.sce new file mode 100755 index 000000000..2dbf8da2f --- /dev/null +++ b/3250/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,25 @@ +clc +// Given that +d= 4 // Depth of cut in mm +f = 0.25 // Feed in mm/stroke +alpha = 10 // Rake angle in degree +shi = 30 // Principal cutting edge angle in Degree +mu =0.6 // Cofficient of friction between chip and tool +T_s = 340 // Ultimate shear stress of cast iron in N/mm^2 +N = 60 // Cutting stroke/min +L = 200 // Length of the job in mm +H = 180 // Hardness of the workpiece in BHN +// Sample Problem 10 on page no. 221 +printf("\n # PROBLEM 4.10 # \n") +lambda = atand(mu) +phi = 45 +alpha-lambda +Fc = f*d*T_s*(cosd(lambda-alpha))/((sind(phi))*(cosd(phi+lambda-alpha))) +Fc_ = Fc*(L/1000) +Wav =Fc_*N/60 +t1 = f*cosd(shi) +U_0 = 0.81 // By using table 4.4 given in the book, In J/mm^3 +Uc = U_0*((t1)^(-.4)) +Q = f*d*L*N/60 +Wav_ = Uc*Q +printf(" \n Avg power consumption = %d W,\n Specific power consumption when hardness of the workpiece is 180 BHN = %d W.",Wav,Wav_) +// Answer in the book for Specific power consumption is given as 294 W diff --git a/3250/CH4/EX4.10/Ex4_10.txt b/3250/CH4/EX4.10/Ex4_10.txt new file mode 100755 index 000000000..b4203e8cb --- /dev/null +++ b/3250/CH4/EX4.10/Ex4_10.txt @@ -0,0 +1,4 @@ + # PROBLEM 4.10 #
+
+ Avg power consumption = 220 W,
+ Specific power consumption when hardness of the workpiece is 180 BHN = 298 W.
\ No newline at end of file diff --git a/3250/CH4/EX4.11/Ex4_11.sce b/3250/CH4/EX4.11/Ex4_11.sce new file mode 100755 index 000000000..23b32d151 --- /dev/null +++ b/3250/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,15 @@ +clc +// Given that +alpha_b = 6 // Back rake angle in Degree +alpha_s = 10 // Side rake angle in Degree +gama = 7 // Front clearance angle in Degree +gama_ = 7 // Side clearance angle in Degree +Shi = 10 // End cutting edge angle in Degree +shi = 30 // Side cutting edge angle in Degree +r= 0.5 // Nose radius in mm +// Sample Problem 11 on page no. 224 +printf("\n # PROBLEM 4.11 # \n") +k = tand(alpha_b) * cosd(shi) - tand(alpha_s) * sind(shi) +printf("\n The value of k=%f,which is near to 0. Hence the case is close to orthogonal one.\n",k) +alpha= atand(((tand(alpha_b) * sind(shi) ) + (tand(alpha_s) * (cosd(shi))))/ (sqrt(1+((tand(alpha_b)*cosd(shi)) - (tand(alpha_s)*sind(shi)))^(2)))) +printf(" \n Normal rake angle = %f°.",alpha) diff --git a/3250/CH4/EX4.11/Ex4_11.txt b/3250/CH4/EX4.11/Ex4_11.txt new file mode 100755 index 000000000..34d742ba0 --- /dev/null +++ b/3250/CH4/EX4.11/Ex4_11.txt @@ -0,0 +1,5 @@ + # PROBLEM 4.11 #
+
+ The value of k=0.002859,which is near to 0. Hence the case is close to orthogonal one.
+
+ Normal rake angle = 11.599142°.
\ No newline at end of file diff --git a/3250/CH4/EX4.12/Ex4_12.sce b/3250/CH4/EX4.12/Ex4_12.sce new file mode 100755 index 000000000..531b6bb73 --- /dev/null +++ b/3250/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,26 @@ +clc +// Given that +alpha_b = 6 // Back rake angle in Degree +alpha_s = 10 // Side rake angle in Degree +gama = 5 // Front clearance angle in Degree +gama_ = 7 // Side clearance angle in Degree +Shi = 10 // End cutting edge angle in Degree +shi = 30 // Side cutting edge angle in Degree +r= 0.55 // Nose radius in mm +d = 2.5 // Depth of cut in mm +f = 0.125 // Feed in mm/revolution +N = 300 // Rpm of the job +T_S = 400 // Ultimate shear stress of the workpiece in N/mm^2 +mu = .6 // Cofficient of the friction between the tool and the chip +// Sample Problem 12 on page no. 225 +printf("\n # PROBLEM 4.12 # \n") +lambda = atand(mu) +alpha= atand(((tand(alpha_b) * sind(shi) ) + (tand(alpha_s) * (cosd(shi))))/ (sqrt(1+((tand(alpha_b)*cosd(shi)) - (tand(alpha_s)*sind(shi)))^(2)))) +phi = 45 + alpha - lambda +t1 = f*cosd(phi) +w = d/cosd(phi) +Fc = w*t1*T_S*(cosd(lambda-alpha))/((sind(phi))*(cosd(phi+lambda-alpha))) +Ft = Fc*tand(lambda-alpha) +Ff = Ft*cosd(shi) +Fr = Ft*sind(shi) +printf(" \n Component of the machining force are as follows -\n Feed force component = % d N, \n Normal thrust force component = % d N.",ceil(Ff),ceil(Fr)) diff --git a/3250/CH4/EX4.12/Ex4_12.txt b/3250/CH4/EX4.12/Ex4_12.txt new file mode 100755 index 000000000..4fe61e0ce --- /dev/null +++ b/3250/CH4/EX4.12/Ex4_12.txt @@ -0,0 +1,6 @@ +
+ # PROBLEM 4.12 #
+
+ Component of the machining force are as follows -
+ Feed force component = 118 N,
+ Normal thrust force component = 68 N.
\ No newline at end of file diff --git a/3250/CH4/EX4.14/Ex4_14.sce b/3250/CH4/EX4.14/Ex4_14.sce new file mode 100755 index 000000000..c1aa294c8 --- /dev/null +++ b/3250/CH4/EX4.14/Ex4_14.sce @@ -0,0 +1,23 @@ +clc +// Given that +D = 20 // Nominal diameter of the drill in mm +T_S = 400 // Shear yield stress of work material in N/mm^2 +N = 240 // Rpm +f = 0.25 // Feed in mm/revolution +mu = 0.6 // Cofficient of friction +// Sample Problem 14 on page no. 230 +printf("\n # PROBLEM 4.14 # \n") +Beta = 118/2 // From the table 4.12 given in the book +shi = 30 // From the table 4.12 given in the book +alpha = atand(((2*(D/4)/(D))*tand(shi))/sind(Beta)) +t1 = (f/2)*sind(Beta) +w = (D/2)/sind(Beta) +lambda = atand(mu) +phi = 45+alpha-lambda +t1 = f/2 +Fc = w*t1*T_S*(cosd(lambda-alpha))/((sind(phi))*(cosd(phi+lambda-alpha))) +Ft = w*t1*T_S*(sind(lambda-alpha))/((sind(phi))*(cosd(phi+lambda-alpha))) +M = .6*Fc*D/1000 +F = 5*Ft*sind(Beta) +printf(" \n The drilling torque = %f N-m, \n Thrust force = %d N.",M,F) +// Answer in the book for drilling torque is given as 18.2 N-m, and for thrust force is given as 1500 N diff --git a/3250/CH4/EX4.14/Ex4_14.txt b/3250/CH4/EX4.14/Ex4_14.txt new file mode 100755 index 000000000..87b1810dd --- /dev/null +++ b/3250/CH4/EX4.14/Ex4_14.txt @@ -0,0 +1,4 @@ + # PROBLEM 4.14 #
+
+ The drilling torque = 17.924639 N-m,
+ Thrust force = 1401 N.
\ No newline at end of file diff --git a/3250/CH4/EX4.15/Ex4_15.sce b/3250/CH4/EX4.15/Ex4_15.sce new file mode 100755 index 000000000..44957246f --- /dev/null +++ b/3250/CH4/EX4.15/Ex4_15.sce @@ -0,0 +1,28 @@ +clc +// Given that +w = 20 // Width of the mild steel block in mm +Z = 20 // No of teeth in milling cutter +D = 50 // Diameter of the milling cutter in mm +alpha = 10 // Radial rake angle in Degree +f = 15 // Feed velocity of the table in mm/min +N =60 // Rpm of the cutter +t = 1 // Depth of cut in mm +mu = 0.5 // Cofficient of friction +T_s = 400 // Shear yield stress in N/mm^2 +t_a = 0.0018 // Avg uncut thickness in mm +// Sample Problem 15 on page no. 235 +printf("\n # PROBLEM 4.15 # \n") +Beta = asind(2*(t/D)) +theta = 2*%pi/Z +t1_max = (2*f/(N*Z))*sqrt(t/D) +lambda = atand(mu) +phi = 45+alpha -lambda +Fc_max = ((w*t1_max*T_s*cosd(lambda-alpha)))/((sind(phi))*(cosd(45))) +T_max = Fc_max*D/(2*1000) +M_av = (1/2)*(Beta*T_max)/theta +omega = 2*%pi*N/60 +U_0 = 1.4 // From the table 4.4 given in the book +Uc_ms = U_0*((t_a)^(-0.4)) +R = f*t*w/60 +U = Uc_ms * R +printf(" \n Power consumption = %f W.",U) diff --git a/3250/CH4/EX4.15/Ex4_15.txt b/3250/CH4/EX4.15/Ex4_15.txt new file mode 100755 index 000000000..2e3d32928 --- /dev/null +++ b/3250/CH4/EX4.15/Ex4_15.txt @@ -0,0 +1,4 @@ +
+ # PROBLEM 4.15 #
+
+ Power consumption = 87.697869 W.
\ No newline at end of file diff --git a/3250/CH4/EX4.16/Ex4_16.sce b/3250/CH4/EX4.16/Ex4_16.sce new file mode 100755 index 000000000..6353972bb --- /dev/null +++ b/3250/CH4/EX4.16/Ex4_16.sce @@ -0,0 +1,29 @@ +clc +// Given that +w = 20 // Width of the mild steel block in mm +Z = 10 // No of teeth in milling cutter +D = 75 // Diameter of the milling cutter in mm +alpha = 10 // Radial rake angle in Degree +f = 100 // Feed velocity of the table in mm/min +N =60 // Rpm of the cutter +t = 5 // Depth of cut in mm +mu = 0.5 // Cofficient of friction +T_s = 400 // Shear yield stress in N/mm^2 +t_a = 0.043 // Avg uncut thickness in mm +// Sample Problem 16 on page no. 238 +printf("\n # PROBLEM 4.16 # \n") +Beta = asind(2*(t/D)) +theta = 2*%pi/Z +t1_max = (2*f/(N*Z))*sqrt(t/D) +lambda = atand(mu) +phi = 45+alpha -lambda +Fc_max = ((w*t1_max*T_s*cosd(lambda-alpha)))/((sind(phi))*(cosd(45))) +T_max = Fc_max*D/(2*1000) +M_av = (1/2)*(Beta*T_max)/theta +omega = 2*%pi*N/60 +U_0 = 1.4 // From the table 4.4 given in the book +Uc_ms = U_0*((t_a)^(-0.4)) +R = f*t*w/60 +U = Uc_ms * R +printf(" \n Power required = %d W.",U) +// Answer in the book for Power required is given as 817 W diff --git a/3250/CH4/EX4.16/Ex4_16.txt b/3250/CH4/EX4.16/Ex4_16.txt new file mode 100755 index 000000000..9c58b5dce --- /dev/null +++ b/3250/CH4/EX4.16/Ex4_16.txt @@ -0,0 +1,3 @@ + # PROBLEM 4.16 #
+
+ Power required = 821 W.
\ No newline at end of file diff --git a/3250/CH4/EX4.17/Ex4_17.sce b/3250/CH4/EX4.17/Ex4_17.sce new file mode 100755 index 000000000..dd1e0a4df --- /dev/null +++ b/3250/CH4/EX4.17/Ex4_17.sce @@ -0,0 +1,20 @@ +clc +// Given that +B = 20 // Width of the cut in mm +Z = 10 // No of teeth in milling cutter +D = 75 // Diameter of the milling cutter in mm +alpha = 10 // Radial rake angle in Degree +f = 25 // Feed velocity of the table in mm/min +N =60 // Rpm of the cutter +t = 5 // Depth of cut in mm +mu = 0.5 // Cofficient of friction +T_s = 400 // Shear yield stress in N/mm^2 +t_a = 0.043 // Avg uncut thickness in mm +// Sample Problem 17 on page no. 240 +printf("\n # PROBLEM 4.17 # \n") +t1_max = 0.01 +lambda = 0.28 // From the table 4.13 Given in the book +nu = 1400 // From the table 4.13 Given in the book +t1_av = t1_max/2 +P = nu*B*t*f*(10^-4)/(6*((t1_av)^(lambda))) +printf(" \n Power required = %f W.",P) diff --git a/3250/CH4/EX4.17/Ex4_17.txt b/3250/CH4/EX4.17/Ex4_17.txt new file mode 100755 index 000000000..655b02783 --- /dev/null +++ b/3250/CH4/EX4.17/Ex4_17.txt @@ -0,0 +1,3 @@ + # PROBLEM 4.17 #
+
+ Power required = 257.161051 W.
\ No newline at end of file diff --git a/3250/CH4/EX4.18/Ex4_18.sce b/3250/CH4/EX4.18/Ex4_18.sce new file mode 100755 index 000000000..b60b4ec4d --- /dev/null +++ b/3250/CH4/EX4.18/Ex4_18.sce @@ -0,0 +1,18 @@ +clc +// Given that +w = 20 // Width of the mild steel block in mm +Z = 10 // No of teeth in milling cutter +D = 75 // Diameter of the milling cutter in mm +alpha = 10 // Radial rake angle in Degree +f = 25 // Feed velocity of the table in mm/min +N =60 // Rpm of the cutter +t = 5 // Depth of cut in mm +mu = 0.5 // Cofficient of friction +T_s = 400 // Shear yield stress in N/mm^2 +t_a = 0.043 // Avg uncut thickness in mm +// Sample Problem 18 on page no. 240 +printf("\n # PROBLEM 4.18 # \n") +R = f*t*w/60 +Uc = 3.3 // Specific energy in J/mm^3 from the table 4.14 Given in the book +U = Uc * R +printf(" \n Power required = %d W.",ceil(U)) diff --git a/3250/CH4/EX4.18/Ex4_18.txt b/3250/CH4/EX4.18/Ex4_18.txt new file mode 100755 index 000000000..c316ee89d --- /dev/null +++ b/3250/CH4/EX4.18/Ex4_18.txt @@ -0,0 +1,4 @@ +
+ # PROBLEM 4.18 #
+
+ Power required = 138 W.
\ No newline at end of file diff --git a/3250/CH4/EX4.19/Ex4_19.sce b/3250/CH4/EX4.19/Ex4_19.sce new file mode 100755 index 000000000..fad6a2c33 --- /dev/null +++ b/3250/CH4/EX4.19/Ex4_19.sce @@ -0,0 +1,18 @@ +clc +// Given that +d = 25 // Diameter of circular hole in mm +t = 20 // Thickness of the steel plate in mm +D = 27 // Enlarged diameter of hole in mm +c= 0.08 // Cut per tooth in mm +alpha = 10 // Radial rake angle in Degree +mu = 0.5 // Cofficient of friction +T_s = 400 // Shear yield stress in N/mm^2 +// Sample Problem 19 on page no. 241 +printf("\n # PROBLEM 4.19 # \n") +lambda=atand(mu) +phi = 45-lambda+alpha +w = %pi*(d+D)/2 +Fc = w*c*T_s*(cosd(lambda-alpha))/((sind(phi))*(cosd(45))) +s = 1.75*sqrt(t) +F = 3*Fc +printf(" \n Peak broaching load = %d N.",ceil(F)) diff --git a/3250/CH4/EX4.19/Ex4_19.txt b/3250/CH4/EX4.19/Ex4_19.txt new file mode 100755 index 000000000..6748f118c --- /dev/null +++ b/3250/CH4/EX4.19/Ex4_19.txt @@ -0,0 +1,3 @@ + # PROBLEM 4.19 #
+
+ Peak broaching load = 22323 N.
\ No newline at end of file diff --git a/3250/CH4/EX4.2/Ex4_2.sce b/3250/CH4/EX4.2/Ex4_2.sce new file mode 100755 index 000000000..d7132e349 --- /dev/null +++ b/3250/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,17 @@ +clc +// Given that +t1 = 0.25 // Undercut thickness in mm +t2 = 0.75 // Chip thickness in mm +w = 2.5 // Width in mm +alpha = 0 // Rake angle in Degree +Fc = 950 // Cutting force in N +Ft = 475 // Thrust force in N +// Sample Problem 2 on page no. 192 +printf("\n # PROBLEM 4.2 # \n") +r = t1/t2 +mu = ((Fc*sind(alpha)) + (Ft*cosd(alpha)))/((Fc*cosd(alpha))-(Ft*sind(alpha))) +phi = atand((r*cosd(alpha))/(1-r*sind(alpha))) +As = t1*w/sind(phi) +Fs = Fc*cosd(phi) - Ft*sind(phi) +T_s = Fs/As +printf("\n Coefficient of the friction between tool and the chip = %f, \n The ultimate shear stress of the material = %f N/mm^2",mu,T_s) diff --git a/3250/CH4/EX4.2/Ex4_2.txt b/3250/CH4/EX4.2/Ex4_2.txt new file mode 100755 index 000000000..08c3a4e75 --- /dev/null +++ b/3250/CH4/EX4.2/Ex4_2.txt @@ -0,0 +1,4 @@ + # PROBLEM 4.2 #
+
+ Coefficient of the friction between tool and the chip = 0.500000,
+ The ultimate shear stress of the material = 380.000000 N/mm^2
\ No newline at end of file diff --git a/3250/CH4/EX4.20/Ex4_20.sce b/3250/CH4/EX4.20/Ex4_20.sce new file mode 100755 index 000000000..0f7abe200 --- /dev/null +++ b/3250/CH4/EX4.20/Ex4_20.sce @@ -0,0 +1,18 @@ +clc +// Given that +D = 250 // Diameter of the wheel in mm +N = 2000 // Rpm of the wheel +f =5 // Plung feed rate in mm/min +C = 3 // Surface density of active grain in mm^-2 +A = 20*15 // Area of mild steel prismatic bar in mm^2 +rg = 15 // In mm^-1 +// Sample Problem 20 on page no. 246 +printf("\n # PROBLEM 4.20 # \n") +t1 = sqrt(f/(%pi*D*N*C*rg)) +U_0 = 1.4 // From the table 4.4 given in the book +Uc= U_0*((t1)^(-.4)) +R = A*f/60 +P = Uc*R +Fc_ = 60000*(P)/(%pi*D*A*C*N) +printf(" \n Power requirement during plunge grinding of the mild steel primatic bar = %d W.",ceil(P)) +// Answer in the book is given as 94 W diff --git a/3250/CH4/EX4.20/Ex4_20.txt b/3250/CH4/EX4.20/Ex4_20.txt new file mode 100755 index 000000000..a8b766869 --- /dev/null +++ b/3250/CH4/EX4.20/Ex4_20.txt @@ -0,0 +1,3 @@ + # PROBLEM 4.20 #
+
+ Power requirement during plunge grinding of the mild steel primatic bar = 943 W.
\ No newline at end of file diff --git a/3250/CH4/EX4.21/Ex4_21.sce b/3250/CH4/EX4.21/Ex4_21.sce new file mode 100755 index 000000000..116ca2f96 --- /dev/null +++ b/3250/CH4/EX4.21/Ex4_21.sce @@ -0,0 +1,19 @@ +clc +// Given that +w = 25 // Width of mild steel block in mm +d= 0.05 // Depth of cut in mm +D = 200 // Diameter of the wheel in mm +N = 3000 // Rpm of the wheel +f =100 // Feed velocity of table in mm/min +C = 3 // No of grits in mm^-2 +rg = 15 // In mm^-1 +// Sample Problem 21 on page no. 248 +printf("\n # PROBLEM 4.21 # \n") +t1_max = sqrt(((6*f)/(%pi*D*N*C*rg))*sqrt(d/D)) +t1_a = t1_max/2 +U_0 = 1.4 // From the table 4.4 given in the book +Uc= U_0*((t1_a)^(-.4)) +R = w*d*f/60 +P = Uc*R +Fc = 60000*(P)/(%pi*D*N) +printf(" \n Grinding force = %d N",Fc) diff --git a/3250/CH4/EX4.21/Ex4_21.txt b/3250/CH4/EX4.21/Ex4_21.txt new file mode 100755 index 000000000..6a0d7bcbe --- /dev/null +++ b/3250/CH4/EX4.21/Ex4_21.txt @@ -0,0 +1,3 @@ + # PROBLEM 4.21 #
+
+ Grinding force = 3 N
\ No newline at end of file diff --git a/3250/CH4/EX4.22/Ex4_22.sce b/3250/CH4/EX4.22/Ex4_22.sce new file mode 100755 index 000000000..5de68f024 --- /dev/null +++ b/3250/CH4/EX4.22/Ex4_22.sce @@ -0,0 +1,14 @@ +clc +// Given that +d= 0.05 // Depth of cut in mm +f =200 // Feed rate in mm/min +theta = 850 // Surface temperature in °C +Theta = 700 // Maximum surface temperature of workpiece surface required to maintain in °C +// Sample Problem 22 on page no. 251 +printf("\n # PROBLEM 4.22 # \n") +K = theta * (f^0.2)/(d^0.9) +r = Theta/K +C = d*f +Dm = (r*C^0.2)^(1/1.1) +fm = C/Dm +printf(" \n Required depth of cut = %f mm,\n Required feed = %d mm/min",Dm,fm) diff --git a/3250/CH4/EX4.22/Ex4_22.txt b/3250/CH4/EX4.22/Ex4_22.txt new file mode 100755 index 000000000..c18cd0a74 --- /dev/null +++ b/3250/CH4/EX4.22/Ex4_22.txt @@ -0,0 +1,4 @@ + # PROBLEM 4.22 #
+
+ Required depth of cut = 0.041910 mm,
+ Required feed = 238 mm/min
\ No newline at end of file diff --git a/3250/CH4/EX4.24/Ex4_24.sce b/3250/CH4/EX4.24/Ex4_24.sce new file mode 100755 index 000000000..ca2b6e9f4 --- /dev/null +++ b/3250/CH4/EX4.24/Ex4_24.sce @@ -0,0 +1,11 @@ +clc +// Given that +shi = 30 // Side cutting edge angle in Degree +lambda = 7 // End cutting edge angle in Degree +r = 0.7 // Nose radius in mm +f = 0.125 // Feed in mm +// Sample Problem 24 on page no. 260 +printf("\n # PROBLEM 4.24 # \n") +H_max = f/(tand(shi)+cotd(lambda)) +H_max_ = (f^2)/(8*r) +printf(" \n Maximum height of uneveness in first tool case = %f mm,\n In second tool case = %f mm",H_max,H_max_) diff --git a/3250/CH4/EX4.24/Ex4_24.txt b/3250/CH4/EX4.24/Ex4_24.txt new file mode 100755 index 000000000..a439e658e --- /dev/null +++ b/3250/CH4/EX4.24/Ex4_24.txt @@ -0,0 +1,5 @@ +
+ # PROBLEM 4.24 #
+
+ Maximum height of uneveness in first tool case = 0.014332 mm,
+ In second tool case = 0.002790 mm
\ No newline at end of file diff --git a/3250/CH4/EX4.25/Ex4_25.sce b/3250/CH4/EX4.25/Ex4_25.sce new file mode 100755 index 000000000..9cd7a56e9 --- /dev/null +++ b/3250/CH4/EX4.25/Ex4_25.sce @@ -0,0 +1,10 @@ +clc +// Given that +Z = 12 // No of teeth +d = 100 // Diameter of cutter in mm +N = 60 // Rpm of cutter +f = 25 // Table feed in mm/min +// Sample Problem 25 on page no. 262 +printf("\n # PROBLEM 4.25 # \n") +H_max = (f^2)/(4*d*(N^2)*(Z^2)) +printf(" \n Maximum height of uneveness = %f mm",H_max) diff --git a/3250/CH4/EX4.25/Ex4_25.txt b/3250/CH4/EX4.25/Ex4_25.txt new file mode 100755 index 000000000..5cb6d7583 --- /dev/null +++ b/3250/CH4/EX4.25/Ex4_25.txt @@ -0,0 +1,4 @@ +
+ # PROBLEM 4.25 #
+
+ Maximum height of uneveness = 0.000003 mm
\ No newline at end of file diff --git a/3250/CH4/EX4.26/Ex4_26.sce b/3250/CH4/EX4.26/Ex4_26.sce new file mode 100755 index 000000000..477c3f53c --- /dev/null +++ b/3250/CH4/EX4.26/Ex4_26.sce @@ -0,0 +1,12 @@ +clc +// Given that +n = 0.25 // Value of exponent of time in Taylor's tool life equation +C = 75 // Value of constant in Taylor's tool life equation +Lc = .15 // Labour cast in $/min +Tc = 2.50 // Total cast of tool in $ +t = 2 // Change time for tool in min +// Sample Problem 26 on page no. 268 +printf("\n # PROBLEM 4.26 # \n") +x = (C)^(1/n) // Where x = k/(f^(1/n)) +v_opt = ((n*x*Lc)/((1-n)*((Lc*t+Tc))))^(n) +printf(" \n Cutting speed that will be lead to minimum cast = %f m/min",v_opt) diff --git a/3250/CH4/EX4.26/Ex4_26.txt b/3250/CH4/EX4.26/Ex4_26.txt new file mode 100755 index 000000000..5411dad46 --- /dev/null +++ b/3250/CH4/EX4.26/Ex4_26.txt @@ -0,0 +1,3 @@ + # PROBLEM 4.26 #
+
+ Cutting speed that will be lead to minimum cast = 27.416642 m/min
\ No newline at end of file diff --git a/3250/CH4/EX4.27/Ex4_27.sce b/3250/CH4/EX4.27/Ex4_27.sce new file mode 100755 index 000000000..7516c28ca --- /dev/null +++ b/3250/CH4/EX4.27/Ex4_27.sce @@ -0,0 +1,24 @@ +clc +// Given that +L = 300 // Length of the bar in mm +d=30 // Diameter of the bar in mm +f_max = 0.25 // Maximum allowable feed in mm/revolution +Lc = .25 // Labour and overhead cast in $/min +Tc = 2 // Regrinding cast in $ +t = 1 // Change time for tool in min +C_X = 2.50 // Cast of tool of material X per piece in $ +C_Y = 3 // Cast of tool of material Y per piece in $ +n_x = 0.1 // Value of exponent of time in Taylor's tool life equation for material X +n_y = 0.16 // Value of exponent of time in Taylor's tool life equation for material Y +C_x = 30 // Value of constant in Taylor's tool life equation for material X +C_y = 76 // Value of constant in Taylor's tool life equation For material Y +// Sample Problem 27 on page no. 269 +printf("\n # PROBLEM 4.27 # \n") +x_x = (C_x)^(1/n_x) // Where x = k/(f^(1/n)) +v_opt_x = ((n_x*x_x*Lc)/((1-n_x)*((Lc*t+Tc))))^(n_x) +Rmin_x = C_X+Lc*t+(Lc*%pi*L*d/(1000*f_max*v_opt_x)) + (Lc*t*(%pi*L*d/(1000*x_x)))*(v_opt_x^(1/n_y))*(v_opt_x^-1)*(f_max^-1)+(Tc*((%pi*L*d/(1000*x_x)))*(v_opt_x^(1/n_x))*(v_opt_x^-1)*(f_max^-1)) +x_y = (C_y)^(1/n_y) // Where x = k/(f^(1/n)) +v_opt_y = ((n_y*x_y*Lc)/((1-n_y)*((Lc*t+Tc))))^(n_y) +Rmin_y = C_Y+Lc*t+(Lc*%pi*L*d/(1000*f_max*v_opt_y)) + (Lc*t*(%pi*L*d/(1000*x_y)))*(v_opt_y^(1/n_y))*(v_opt_y^-1)*(f_max^-1)+(Tc*((%pi*L*d/(1000*x_y)))*(v_opt_y^(1/n_y))*(v_opt_y^-1)*(f_max^-1)) +printf(" \n The minimum cast per piece\n When material X is used = %f $,\n When material Y is used = %f $",Rmin_x,Rmin_y) +printf("\n So material Y will be suitable for tool as it has low cast") diff --git a/3250/CH4/EX4.27/Ex4_27.txt b/3250/CH4/EX4.27/Ex4_27.txt new file mode 100755 index 000000000..38e0fc2ae --- /dev/null +++ b/3250/CH4/EX4.27/Ex4_27.txt @@ -0,0 +1,7 @@ +
+ # PROBLEM 4.27 #
+
+ The minimum cast per piece
+ When material X is used = 4.357033 $,
+ When material Y is used = 4.070737 $
+ So material Y will be suitable for tool as it has low cast
\ No newline at end of file diff --git a/3250/CH4/EX4.28/Ex4_28.sce b/3250/CH4/EX4.28/Ex4_28.sce new file mode 100755 index 000000000..f9f2f8e8a --- /dev/null +++ b/3250/CH4/EX4.28/Ex4_28.sce @@ -0,0 +1,12 @@ +clc +// Given that +n = 0.25 // Value of exponent of time in Taylor's tool life equation +C = 75 // Value of constant in Taylor's tool life equation +Lc = .15 // Labour cast in $/min +Tc = 2.50 // Total cast of tool in $ +t = 2 // Change time for tool in min +// Sample Problem 28 on page no. 271 +printf("\n # PROBLEM 4.28 # \n") +x = (C)^(1/n) // Where x = k/(f^(1/n)) +v_opt = ((n*x)/((1-n)*t))^(n) +printf(" \n Optimum cutting speed for maximum production rate for the job = %f m/min",v_opt) diff --git a/3250/CH4/EX4.28/Ex4_28.txt b/3250/CH4/EX4.28/Ex4_28.txt new file mode 100755 index 000000000..1b92aa9e6 --- /dev/null +++ b/3250/CH4/EX4.28/Ex4_28.txt @@ -0,0 +1,3 @@ + # PROBLEM 4.28 #
+
+ Optimum cutting speed for maximum production rate for the job = 47.920733 m/min
\ No newline at end of file diff --git a/3250/CH4/EX4.3/Ex4_3.sce b/3250/CH4/EX4.3/Ex4_3.sce new file mode 100755 index 000000000..3ac6506b9 --- /dev/null +++ b/3250/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,18 @@ +clc +// Given that +alpha = 10 // Rake angle of tool in Degree +v = 200 // Cutting speed in m/min +t1 = 0.2 // Uncut thickness in mm +w = 2 // Width of cut in mm +mu = 0.5 // Avg value of the cofficient of tbe friction +T_S = 400 // Shear stress of the work material in N/mm^2 +// Sample Problem 3 on page no. 193 +printf("\n # PROBLEM 4.3 # \n") +lambda = atand(mu) +phi = (90 + alpha - lambda)/2 +Fs = (w*t1*T_S)/(sind(phi)) +R = Fs/(cosd(phi+lambda-alpha)) +Fc = R*(cosd(lambda-alpha)) +Ft = R*(sind(lambda-alpha)) +printf("\n Shear angle = %f°, \n Cutting force = %d N, \n Thrust force = %d N,",phi,Fc,Ft) +// Answer in the book for cutting force is given as 420 N and for thrust force is given as 125 N diff --git a/3250/CH4/EX4.3/Ex4_3.txt b/3250/CH4/EX4.3/Ex4_3.txt new file mode 100755 index 000000000..7ab155fda --- /dev/null +++ b/3250/CH4/EX4.3/Ex4_3.txt @@ -0,0 +1,5 @@ + # PROBLEM 4.3 #
+
+ Shear angle = 36.717474°,
+ Cutting force = 429 N,
+ Thrust force = 127 N,
\ No newline at end of file diff --git a/3250/CH4/EX4.4/Ex4_4.sce b/3250/CH4/EX4.4/Ex4_4.sce new file mode 100755 index 000000000..319c7254a --- /dev/null +++ b/3250/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,25 @@ +clc +// Given that +alpha = 10 // Rake angle of tool in Degree +v = 200 // Cutting speed in m/min +t1 = 0.2 // Uncut thickness in mm +w = 2 // Width of cut in mm +mu = 0.5 // Avg value of the cofficient of tbe friction +T_S = 400 // Shear stress of the work material in N/mm^2 +Cm = 70 // Machining constant in Degree +// Sample Problem 4 on page no. 194 +printf("\n # PROBLEM 4.4 # \n") +lambda = atand(mu) +phi = (Cm + alpha - lambda)/2 +Fs = (w*t1*T_S)/(sind(phi)) +R = Fs/(cosd(phi+lambda-alpha)) +Fc = R*(cosd(lambda-alpha)) +Ft = R*(sind(lambda-alpha)) +// Using Lee and Shaffer relation +phi_ = 45-lambda+alpha +Fs_ = (w*t1*T_S)/(sind(phi_)) +R_ = Fs_/(cosd(phi_+lambda-alpha)) +Fc_ = R_*(cosd(lambda-alpha)) +Ft_ = R_*(sind(lambda-alpha)) +printf("\n Shear angle = %f°, \n Cutting force = %f N, \n Thrust force = %f N \n Using Lee and Shaffer relation- \n Shear angle = %f°, \n Cutting force = %f N, \n Thrust force = %f N,",phi,Fc,Ft,phi_,Fc_,Ft_) +// Answer in the book for cutting force is given as 486.9 N and for thrust force is given as 144.9 N , When using Lee and Shaffer relation answer in the book for cutting force is given as 481.9 N and for trust force is given as 160.6 N diff --git a/3250/CH4/EX4.4/Ex4_4.txt b/3250/CH4/EX4.4/Ex4_4.txt new file mode 100755 index 000000000..42f6574c5 --- /dev/null +++ b/3250/CH4/EX4.4/Ex4_4.txt @@ -0,0 +1,9 @@ + # PROBLEM 4.4 #
+
+ Shear angle = 26.717474°,
+ Cutting force = 468.567296 N,
+ Thrust force = 139.374821 N
+ Using Lee and Shaffer relation-
+ Shear angle = 28.434949°,
+ Cutting force = 455.482874 N,
+ Thrust force = 135.482874 N,
\ No newline at end of file diff --git a/3250/CH4/EX4.5/Ex4_5.sce b/3250/CH4/EX4.5/Ex4_5.sce new file mode 100755 index 000000000..497435ea5 --- /dev/null +++ b/3250/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,16 @@ +clc +// Given that +t1 = 0.25 // Uncut thickness in mm +w = 2.5 // Width of cut in mm +U_0 = 1.4 // In J/mm^3 +alpha = 0 // Rake angle in degree +mu = 0.5 // Cofficient of the friction +T_s = 400 // Shear stress in N/mm^2 +// Sample Problem 5 on page no. 196 +printf("\n # PROBLEM 4.5 # \n") +lambda = atand(mu) +Fc = 1000*(t1*w*U_0)*((t1)^(-.4)) +phi = 45 + alpha - atand(mu) +Fc_ = (w*t1*T_s*cosd(lambda-alpha))/((sind(phi)) *cosd(phi+lambda-alpha)) +printf(" \n The order of magnitude of cutting force = %d N,\n Using Lee and Shaffer relation- \n The order of magnitude of cutting force = %d N.",Fc,Fc_) +// Answer in the book for cutting force is given as 1517 N diff --git a/3250/CH4/EX4.5/Ex4_5.txt b/3250/CH4/EX4.5/Ex4_5.txt new file mode 100755 index 000000000..86c03f189 --- /dev/null +++ b/3250/CH4/EX4.5/Ex4_5.txt @@ -0,0 +1,5 @@ + # PROBLEM 4.5 #
+
+ The order of magnitude of cutting force = 1523 N,
+ Using Lee and Shaffer relation-
+ The order of magnitude of cutting force = 1000 N.
\ No newline at end of file diff --git a/3250/CH4/EX4.6/Ex4_6.sce b/3250/CH4/EX4.6/Ex4_6.sce new file mode 100755 index 000000000..5611c2b36 --- /dev/null +++ b/3250/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,32 @@ +clc +// Given that +v = 2 // Cutting speed in m/sec +D = 7200 // Density of mild steel in kg /m^3 +k = 43.6 // Thermal conductivity in W/m-°c +c = 502 // Specific heat of the material in J/kg-°c +t1 = 0.25 // Uncut thickness in mm +w =2 // Width of cut in mm +theta_0 = 40 // Initial temp of the workpiece in Degree +alpha = 0 // Rake angle in degree +mu = 0.5 // Cofficient of the friction +T_s = 400e6 // Shear stress in N/m^2 +// Sample Problem 6 on page no. 199 +printf("\n # PROBLEM 4.6 # \n") +lambda = atand(mu) +phi = 45 + alpha - lambda +Fs = (w*t1*T_s)*(10^-6)/(sind(phi)) +R = Fs / (cosd(phi+lambda-alpha)) +Fc = R *(cosd(lambda-alpha)) +r = sind(phi)/(cosd(phi-alpha)) +Ft= Fc *(tand(lambda - alpha)) +F = Fc *(sind(alpha))+Ft*(cosd(alpha)) +Ws = F*r*v +Wp = Fc*v-F*r*v +zeta = D*c*v*t1*(10^-3)/k +zeta_ = zeta*tand(phi) +nu = 0.15 *(log(27.5/(zeta_))) +theta_P = (1-nu)*Wp/(D*c*v*t1*w*(10^-6)) +theta_S = 1.13 *(sqrt(1/(D*c*v*t1*(10^-3)*k*(1+tand(phi-alpha)))))*(Ws/w)*(10^3) +theta = theta_0+theta_S+ theta_P +printf(" \n Maximum temperature along the rake face of the tool = %d°C.",theta) +// Answer in the book is given as 823°C diff --git a/3250/CH4/EX4.6/Ex4_6.txt b/3250/CH4/EX4.6/Ex4_6.txt new file mode 100755 index 000000000..f58c28a3e --- /dev/null +++ b/3250/CH4/EX4.6/Ex4_6.txt @@ -0,0 +1,3 @@ + # PROBLEM 4.6 #
+
+ Maximum temperature along the rake face of the tool = 835°C.
\ No newline at end of file diff --git a/3250/CH4/EX4.7/Ex4_7.sce b/3250/CH4/EX4.7/Ex4_7.sce new file mode 100755 index 000000000..23d8e49ce --- /dev/null +++ b/3250/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,34 @@ +clc +// Given that +theta_ = 40 //Ambient temperature in°C +v = 2 // Cutting speed in m/sec +D = 7200 // Density of mild steel in kg /m^3 +k = 43.6 // Thermal conductivity in W/m-°c +c = 502 // Specific heat of the material in J/kg-°c +t1 = 0.25 // Uncut thickness in mm +w =2 // Width of cut in mm +alpha = 0 // Rake angle in degree +mu = 0.5 // Cofficient of the friction +T_s = 400e6 // Shear stress in N/m^2 +H = 350 // Hardness of SAE 1040 steel in HV(Vicker hardness) +// Sample Problem 7 on page no. 206 +printf("\n # PROBLEM 4.7 # \n") +lambda = atand(mu) +phi = 45 + alpha - lambda +Fs = (w*t1*T_s)*(10^-6)/(sind(phi)) +R = Fs / (cosd(phi+lambda-alpha)) +Fc = R *(cosd(lambda-alpha)) +r = sind(phi)/(cosd(phi-alpha)) +Ft= Fc *(tand(lambda - alpha)) +F = Fc *(sind(alpha))+Ft*(cosd(alpha)) +Ws = F*r*v +Wp = Fc*v-F*r*v +zeta = D*c*v*t1*(10^-3)/k +zeta_ = zeta*tand(phi) +nu = 0.15 *(log(27.5/(zeta_))) +Theta_0v = ((1-nu)*Wp + Ws)/ (D*c*v*t1*w*(10^-6)) +H_ = 1.5 *(H) +theta_lim = 700*((1-(H_/850))^(1/3.1)) +v_lim = (theta_lim/309)^(1/0.5) +printf(" \n Maximum speed at which cutting is passible = %f m/sec.",v_lim) + diff --git a/3250/CH4/EX4.7/Ex4_7.txt b/3250/CH4/EX4.7/Ex4_7.txt new file mode 100755 index 000000000..9e3d0dab5 --- /dev/null +++ b/3250/CH4/EX4.7/Ex4_7.txt @@ -0,0 +1,3 @@ + # PROBLEM 4.7 #
+
+ Maximum speed at which cutting is passible = 2.759951 m/sec.
\ No newline at end of file diff --git a/3250/CH4/EX4.8/Ex4_8.sce b/3250/CH4/EX4.8/Ex4_8.sce new file mode 100755 index 000000000..7b3793db1 --- /dev/null +++ b/3250/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,11 @@ +clc +// Given that +alpha = 0 // Rake angle in degree +gama = 3 // Clearance angle in Degree +w = 1 // Maximum length of flank wear allowed in mm +gama_ = 7 // Increased clearance angle in Degree +// Sample Problem 8 on page no. 212 +printf("\n # PROBLEM 4.8 # \n") +I_per = (((tand(gama_))-(tand(gama)))/tand(gama))*100 +printf(" \n Percentage increase in tool life = %d percent.",I_per) + diff --git a/3250/CH4/EX4.8/Ex4_8.txt b/3250/CH4/EX4.8/Ex4_8.txt new file mode 100755 index 000000000..15b9d8ba9 --- /dev/null +++ b/3250/CH4/EX4.8/Ex4_8.txt @@ -0,0 +1,3 @@ + # PROBLEM 4.8 #
+
+ Percentage increase in tool life = 134 percent.
\ No newline at end of file diff --git a/3250/CH4/EX4.9/Ex4_9.sce b/3250/CH4/EX4.9/Ex4_9.sce new file mode 100755 index 000000000..7b912e9d2 --- /dev/null +++ b/3250/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,18 @@ +clc +// Given that +d= 4 // Depth of cut in mm +f = 0.25 // Feed in mm/stroke +alpha = 10 // Rake angle in degree +shi = 30 // Principal cutting edge angle in Degree +mu =0.6 // Cofficient of friction between chip and tool +T_s = 340 // Ultimate shear stress of cast iron in N/mm^2 +// Sample Problem 9 on page no. 220 +printf("\n # PROBLEM 4.9 # \n") +lambda = atand(mu) +phi = 45 +alpha-lambda +Fc = f*d*T_s*(cosd(lambda-alpha))/((sind(phi))*(cosd(phi+lambda-alpha))) +Ft = Fc*(sind(lambda-alpha))/(cosd(lambda-alpha)) +Ff = Ft*(cosd(shi)) +Fn = Ft*(sind(shi)) +printf(" \n The three components of machinig force are as follows-\n Thrust force = %d N,\n Feed force component = %d N,\n Normal thrust force component = %d N.",Ft,Ff,Fn) + diff --git a/3250/CH4/EX4.9/Ex4_9.txt b/3250/CH4/EX4.9/Ex4_9.txt new file mode 100755 index 000000000..26760b0fb --- /dev/null +++ b/3250/CH4/EX4.9/Ex4_9.txt @@ -0,0 +1,6 @@ + # PROBLEM 4.9 #
+
+ The three components of machinig force are as follows-
+ Thrust force = 422 N,
+ Feed force component = 365 N,
+ Normal thrust force component = 211 N.
\ No newline at end of file diff --git a/3250/CH5/EX5.1/Ex5_1.sce b/3250/CH5/EX5.1/Ex5_1.sce new file mode 100755 index 000000000..fcfef018b --- /dev/null +++ b/3250/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,16 @@ +clc +// Given that +A = 20 // Value of A in voltage length characteristic equation +B = 40 // Value of B in voltage length characteristic equation +v= 80 // Open circuit voltage in V +I = 1000 // Short circuit current in amp +// Sample Problem 1 on page no. 285 +printf("\n # PROBLEM 5.1 # \n") +l=poly(0,"l") +i = ((v-A)-(B* l))*(I/v) +V = (A+B*l)// Given in the question +P = V*i +k = derivat(P) +L=roots(k) +Pmax=((v-A)-(B* L))*(I/v)*(A+B*L) +printf("\n Maximum power of the arc = %d kVA",Pmax/1000) diff --git a/3250/CH5/EX5.1/Ex5_1.txt b/3250/CH5/EX5.1/Ex5_1.txt new file mode 100755 index 000000000..848d6e537 --- /dev/null +++ b/3250/CH5/EX5.1/Ex5_1.txt @@ -0,0 +1,3 @@ + # PROBLEM 5.1 #
+
+ Maximum power of the arc = 20 kVA
\ No newline at end of file diff --git a/3250/CH5/EX5.2/Ex5_2.sce b/3250/CH5/EX5.2/Ex5_2.sce new file mode 100755 index 000000000..819ffb42e --- /dev/null +++ b/3250/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,12 @@ +clc +// Given that +N =25 // No. of bridges per cm^2 +r = 0.1 // Radius of bridge in mm +rho = 2e-5 // Resistivity of the material in ohm-cm +v= 5 // Applied voltage in V +// Sample Problem 2 on page no. 288 +printf("\n # PROBLEM 5.2 # \n") +Rc = 0.85*rho/(N*%pi*r*0.1) +Q = (v^2)/Rc +printf("\n Rate of heat generated per unit area = %e W/cm^2",Q) +// Answer in the book is given as 1.136e5 W/cm^2 diff --git a/3250/CH5/EX5.2/Ex5_2.txt b/3250/CH5/EX5.2/Ex5_2.txt new file mode 100755 index 000000000..52934c015 --- /dev/null +++ b/3250/CH5/EX5.2/Ex5_2.txt @@ -0,0 +1,4 @@ +
+ # PROBLEM 5.2 #
+
+ Rate of heat generated per unit area = 1.154997e+06 W/cm^2
\ No newline at end of file diff --git a/3250/CH5/EX5.3/Ex5_3.sce b/3250/CH5/EX5.3/Ex5_3.sce new file mode 100755 index 000000000..9a0c02119 --- /dev/null +++ b/3250/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,19 @@ +clc +// Given that +P = 2.5 // Power in kVA +t = 3 // Thickness of steel plate in mm +T = 85 // Percentage of total time when arc is on +alpha = 1.2e-5 // Thermal diffusivity of steel in m^2/sec +k = 43.6 // Thermal conductivity of steel in W/m-°C +theta_ = 1530 // Melting point of steel in °C +theta = 30 // Ambient temperature in °C +gama = 60 // Angle in degree +// Sample Problem on page no. 292 +printf("\n # PROBLEM 5.3 # \n") +C = T/100 +Q = C*P*10^3 +w = t/sind(gama) +theta_m = theta_ - theta +v_max = (4*alpha/(w*(10^-3)))*((Q/(8*k*theta_m*t*(10^-3)))-0.2) +printf("\n Maximum passible welding speed = %f m/sec",v_max) +// Answer in the book is given as 0.0146 m/sec diff --git a/3250/CH5/EX5.3/Ex5_3.txt b/3250/CH5/EX5.3/Ex5_3.txt new file mode 100755 index 000000000..e39d3a74a --- /dev/null +++ b/3250/CH5/EX5.3/Ex5_3.txt @@ -0,0 +1,3 @@ + # PROBLEM 5.3 #
+
+ Maximum passible welding speed = 0.015988 m/sec
\ No newline at end of file diff --git a/3250/CH5/EX5.4/Ex5_4.sce b/3250/CH5/EX5.4/Ex5_4.sce new file mode 100755 index 000000000..3a7c296fa --- /dev/null +++ b/3250/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,14 @@ +clc +// Given that +t = 1.2 // Thickness of aluminium sheet in mm +t_ = 0.25 // Adhesive thickness in mm +l = 12 // Overlapped length in mm +E = 703 // Modulus of elastisity in N/mm^2 +G = 11.9 // Shear modulus of adhesive in N/mm^2 +T_S = 0.6 // Ultimate shear stress in N/mm^2 +// Sample Problem 4 on page no. 303 +printf("\n # PROBLEM 5.4 # \n") +K = (((l^2)*G)/(2*E*t*t_))^(1/2) +T = T_S/K +printf("\n The maximum shear stress the lap joint can withstand = %f N/mm^2",T) +// Answer in the book is given as 0.274 N/mm^2 diff --git a/3250/CH5/EX5.4/Ex5_4.txt b/3250/CH5/EX5.4/Ex5_4.txt new file mode 100755 index 000000000..3d9bd9d25 --- /dev/null +++ b/3250/CH5/EX5.4/Ex5_4.txt @@ -0,0 +1,3 @@ + # PROBLEM 5.4 #
+
+ The maximum shear stress the lap joint can withstand = 0.297680 N/mm^2
\ No newline at end of file diff --git a/3250/CH6/EX6.1/Ex6_1.sce b/3250/CH6/EX6.1/Ex6_1.sce new file mode 100755 index 000000000..1a6e726b9 --- /dev/null +++ b/3250/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,19 @@ +clc +// Given that +a = 5 // Side of the square hole in mm +t = 4 // Thickness of tungsten plate in mm +d = 0.01 // Diameter of abraisive grains in mm +F = 3.5 // Force for feeding in N +A =25e-3 // Amplitude of tool oscillation in mm +f = 25e3 // Frequency in Hz +Hw = 6900 // Fracture hardness of WC in N/mm^2 +// Sample Problem 1 on page no. 332 +printf("\n # PROBLEM 6.1 # \n") +Z = (1/2)*(4*s^2)/(%pi*d^2) +lambda = 5 +d1 = (d^2) +h_w = (sqrt((8*F*A)/(%pi*Z*d1*Hw*(1+lambda)))) +Q = (2/3)*((d1*h_w)^(3/2))*Z*f*%pi +t = (a^2)*t/(Q*60) +printf("\n The approximate time required = %f min",t) +// Answer in the book is given as 13.66 min diff --git a/3250/CH6/EX6.1/Ex6_1.txt b/3250/CH6/EX6.1/Ex6_1.txt new file mode 100755 index 000000000..b62fd3a6a --- /dev/null +++ b/3250/CH6/EX6.1/Ex6_1.txt @@ -0,0 +1,3 @@ + # PROBLEM 6.1 #
+
+ The approximate time required = 14.262125 min
\ No newline at end of file diff --git a/3250/CH6/EX6.10/Ex6_10.sce b/3250/CH6/EX6.10/Ex6_10.sce new file mode 100755 index 000000000..7bef113d3 --- /dev/null +++ b/3250/CH6/EX6.10/Ex6_10.sce @@ -0,0 +1,15 @@ +clc +// Given that +R = 50 // Resistance in relaxation circuit in ohm +C = 10 // Capacitance in relaxation circuit in micro F +V = 200 // Supply voltage in Volt +V_ = 150 // Minimum required voltage for discharge in Volt +// Sample Problem 10 on page no. 382 +printf("\n # PROBLEM 6.10 # \n") +E = (1/2)*C*(10^-6)*(V_^2) +tc = R*C*(10^-6)*log(V/(V-V_)) +W = (E/tc)*(10^-3) +Q = 27.4*(W^(1.54)) +Hrms = 1.11*(Q^0.384) +printf("\n Surface roughness = %f micro meter",Hrms) +// Answer in the book is given as 5.16 micro meter diff --git a/3250/CH6/EX6.10/Ex6_10.txt b/3250/CH6/EX6.10/Ex6_10.txt new file mode 100755 index 000000000..2eecca065 --- /dev/null +++ b/3250/CH6/EX6.10/Ex6_10.txt @@ -0,0 +1,3 @@ + # PROBLEM 6.10 #
+
+ Surface roughness = 1.350325 micro meter
\ No newline at end of file diff --git a/3250/CH6/EX6.11/Ex6_11.sce b/3250/CH6/EX6.11/Ex6_11.sce new file mode 100755 index 000000000..15b8ec5b0 --- /dev/null +++ b/3250/CH6/EX6.11/Ex6_11.sce @@ -0,0 +1,10 @@ +clc +// Given that +w = 150 // Width of slot in micro meter +t = 1 // Thickness of tungsten sheet in mm +P = 5 // Power of electron beam in KW +// Sample Problem 11 on page no. 391 +printf("\n # PROBLEM 6.11 # \n") +C = 12 // Specific power consumption for tugsten in W/(mm^3/min) from the table 6.7 given in the book +v = (P*(1000)/C)*(1000/(w*t))*(1/600) +printf("\n Speed of cutting = %f cm/sec",v) diff --git a/3250/CH6/EX6.11/Ex6_11.txt b/3250/CH6/EX6.11/Ex6_11.txt new file mode 100755 index 000000000..0ff9c85cb --- /dev/null +++ b/3250/CH6/EX6.11/Ex6_11.txt @@ -0,0 +1,3 @@ + # PROBLEM 6.11 #
+
+ Speed of cutting = 4.629630 cm/sec
\ No newline at end of file diff --git a/3250/CH6/EX6.12/Ex6_12.sce b/3250/CH6/EX6.12/Ex6_12.sce new file mode 100755 index 000000000..a0c7dc606 --- /dev/null +++ b/3250/CH6/EX6.12/Ex6_12.sce @@ -0,0 +1,8 @@ +clc +// Given that +V = 150e3 // Acceleration voltage in V +// Sample Problem 12 on page no. 392 +printf("\n # PROBLEM 6.12 # \n") +D = 76e-7 // Density of steel in kg/mm^3 +Delta = 2.6*(10^-17)*((V^2)/D) +printf("\n Electron range = %d micro meter",ceil(Delta*(10^3))) diff --git a/3250/CH6/EX6.12/Ex6_12.txt b/3250/CH6/EX6.12/Ex6_12.txt new file mode 100755 index 000000000..bce5a1db7 --- /dev/null +++ b/3250/CH6/EX6.12/Ex6_12.txt @@ -0,0 +1,3 @@ + # PROBLEM 6.12 #
+
+ Electron range = 77 micro meter
\ No newline at end of file diff --git a/3250/CH6/EX6.13/Ex6_13.sce b/3250/CH6/EX6.13/Ex6_13.sce new file mode 100755 index 000000000..029ca371d --- /dev/null +++ b/3250/CH6/EX6.13/Ex6_13.sce @@ -0,0 +1,13 @@ +clc +// Given that +w = 0.015 // Width of slot in cm +t = 1 // Thickness of tungsten sheet in mm +P = 5e3 // Power of electron beam in W +// Sample Problem 13 on page no. 395 +printf("\n # PROBLEM 6.13 # \n") +rho_c = 2.71 // Value of volume specific heat for tugsten in J/cm^3 +k = 2.15 // Thermal conductivity of tungsten in W/cm-°C +T_m = 3400 // Melting temperture in °C +Z = t/10 // In cm +v = (0.1^2)*(P^2)/((T_m^2)*(Z^2)*(k*w*rho_c)) +printf("\n Speed of cutting = %f cm/sec",v) diff --git a/3250/CH6/EX6.13/Ex6_13.txt b/3250/CH6/EX6.13/Ex6_13.txt new file mode 100755 index 000000000..78cd2ff80 --- /dev/null +++ b/3250/CH6/EX6.13/Ex6_13.txt @@ -0,0 +1,4 @@ +
+ # PROBLEM 6.13 #
+
+ Speed of cutting = 24.744755 cm/sec
\ No newline at end of file diff --git a/3250/CH6/EX6.14/Ex6_14.sce b/3250/CH6/EX6.14/Ex6_14.sce new file mode 100755 index 000000000..f620fbb45 --- /dev/null +++ b/3250/CH6/EX6.14/Ex6_14.sce @@ -0,0 +1,13 @@ +clc +// Given that +I = 1e5 // Power intensity of laser beam in W/mm^2 +T_m = 3400 // Melting temperture of tungsten in °C +rho_c = 2.71 // Value of volume specific heat for tugsten in J/cm^3 +k = 2.15 // Thermal conductivity of tungsten in W/cm-°C +p_a = 10 // Percentage of beam absorbed +// Sample Problem 14 on page no. 399 +printf("\n # PROBLEM 6.14 # \n") +alpha = k/rho_c +H = (p_a/100)*(I)*(100) +tm = (%pi/alpha)*((T_m*k)/(2*H))^(2) +printf("\n Time required for the surface to reach the melting point = %f sec",tm) diff --git a/3250/CH6/EX6.14/Ex6_14.txt b/3250/CH6/EX6.14/Ex6_14.txt new file mode 100755 index 000000000..d36828dd3 --- /dev/null +++ b/3250/CH6/EX6.14/Ex6_14.txt @@ -0,0 +1,4 @@ +
+ # PROBLEM 6.14 #
+
+ Time required for the surface to reach the melting point = 0.000053 sec
\ No newline at end of file diff --git a/3250/CH6/EX6.15/Ex6_15.sce b/3250/CH6/EX6.15/Ex6_15.sce new file mode 100755 index 000000000..a8a3a82a5 --- /dev/null +++ b/3250/CH6/EX6.15/Ex6_15.sce @@ -0,0 +1,17 @@ +clc +// Given that +I = 1e5 // Power intensity of laser beam in W/mm^2 +d = 200 // Focused diameter of incident beam in micro meter +T_m = 3400 // Melting temperture of tungsten in °C +rho_c = 2.71 // Value of volume specific heat for tugsten in J/cm^3 +k = 2.15 // Thermal conductivity of tungsten in W/cm-°C +p_a = 10 // Percentage of beam absorbed +// Sample Problem 15 on page no. 400 +printf("\n # PROBLEM 6.15 # \n") +H = (p_a/100)*(I)*(100) +alpha = k/rho_c +zeta = 0.5 // Fr0m the standard table + // By solving the equation T_m = ((2*H)*(sqrt(alpha*tm))/k)*((1/sqrt(%pi))-ierfc(d/(4*sqrt(alpha*tm)))) +tm = 1/((200^2)*(zeta^2)*(alpha)) +printf("\n Time required for the centre of the circular spot to reach the melting point = %f sec",tm) +// Answer in the book is given as 0.00013 sec diff --git a/3250/CH6/EX6.15/Ex6_15.txt b/3250/CH6/EX6.15/Ex6_15.txt new file mode 100755 index 000000000..59a689d08 --- /dev/null +++ b/3250/CH6/EX6.15/Ex6_15.txt @@ -0,0 +1,4 @@ +
+ # PROBLEM 6.15 #
+
+ Time required for the centre of the circular spot to reach the melting point = 0.000126 sec
\ No newline at end of file diff --git a/3250/CH6/EX6.16/Ex6_16.sce b/3250/CH6/EX6.16/Ex6_16.sce new file mode 100755 index 000000000..c4b285d42 --- /dev/null +++ b/3250/CH6/EX6.16/Ex6_16.sce @@ -0,0 +1,11 @@ +clc +// Given that +d = 200 // Diameter of focussed laser beam in micro meter +T_m = 3400 // Melting temperture of tungsten in °C +k = 2.15 // Thermal conductivity of tungsten in W/cm-°C +p_a = 10 // Percentage of beam absorbed +// Sample Problem 16 on page no. 401 +printf("\n # PROBLEM 6.16 # \n") +H = 2*k*T_m/(d*10^-4) +I = H/(p_a/100) +printf("\n Minimum value of beam power intensity to achieve the melting = %e W/cm^2",I) diff --git a/3250/CH6/EX6.16/Ex6_16.txt b/3250/CH6/EX6.16/Ex6_16.txt new file mode 100755 index 000000000..a8ca910d6 --- /dev/null +++ b/3250/CH6/EX6.16/Ex6_16.txt @@ -0,0 +1,3 @@ + # PROBLEM 6.16 #
+
+ Minimum value of beam power intensity to achieve the melting = 7.310000e+06 W/cm^2
\ No newline at end of file diff --git a/3250/CH6/EX6.17/Ex6_17.sce b/3250/CH6/EX6.17/Ex6_17.sce new file mode 100755 index 000000000..8d05b866f --- /dev/null +++ b/3250/CH6/EX6.17/Ex6_17.sce @@ -0,0 +1,15 @@ +clc +// Given that +I = 1e5 // Power intensity of laser beam in W/mm^2 +t = 0.5 // Thickness of tungsten sheet in mm +d = 200 // Drill diameter in micro meter +P = 3e4 // Energy required per unit volume to vapourize tungsten in J/cm^3 +p_e = 10 // Percentage efficiency +T_m = 3400 // Melting temperture of tungsten in °C +k = 2.15 // Thermal conductivity of tungsten in W/cm-°C +// Sample Problem 17 on page no. 403 +printf("\n # PROBLEM 6.17 # \n") +H = (p_e/100)*(I)*(100) +v = H/P +T = t*(0.1)/(v) +printf("\n The time required to drill a through hole = %f sec",T) diff --git a/3250/CH6/EX6.17/Ex6_17.txt b/3250/CH6/EX6.17/Ex6_17.txt new file mode 100755 index 000000000..0aaed52c3 --- /dev/null +++ b/3250/CH6/EX6.17/Ex6_17.txt @@ -0,0 +1,4 @@ +
+ # PROBLEM 6.17 #
+
+ The time required to drill a through hole = 0.001500 sec
\ No newline at end of file diff --git a/3250/CH6/EX6.2/Ex6_2.sce b/3250/CH6/EX6.2/Ex6_2.sce new file mode 100755 index 000000000..140cb91b4 --- /dev/null +++ b/3250/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,10 @@ +clc +// Given that +r = 1/3 // Ratio of hardness values of copper and steel +// Sample Problem 2 on page no. 335 +printf("\n # PROBLEM 6.2 # \n") +R_Q = (r)^(3/4) +R_t = 1/R_Q +P_R = (1-(1/R_t))*100 +printf("\n Percentage change in cutting time when tool is changed from coppper to steel = %d percent(reduction)",P_R) + diff --git a/3250/CH6/EX6.2/Ex6_2.txt b/3250/CH6/EX6.2/Ex6_2.txt new file mode 100755 index 000000000..31a63df76 --- /dev/null +++ b/3250/CH6/EX6.2/Ex6_2.txt @@ -0,0 +1,3 @@ + # PROBLEM 6.2 #
+
+ Percentage change in cutting time when tool is changed from coppper to steel = 56 percent(reduction)
\ No newline at end of file diff --git a/3250/CH6/EX6.3/Ex6_3.sce b/3250/CH6/EX6.3/Ex6_3.sce new file mode 100755 index 000000000..7a2270c68 --- /dev/null +++ b/3250/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,11 @@ +clc +// Given that +m = 5 // Romoval rate in cm^3/min +A = 56 // Atomic gram weight in gm +Z = 2 // Valence at which dissolation takes place +D = 7.8 // Density of iron in gm/cm^3 +// Sample Problem 3 on page no. 345 +printf("\n # PROBLEM 6.3 # \n") +I = (m/60)*(D*Z*96500)/(A) +printf("\n Current required = %d amp",I) + diff --git a/3250/CH6/EX6.3/Ex6_3.txt b/3250/CH6/EX6.3/Ex6_3.txt new file mode 100755 index 000000000..1df49a317 --- /dev/null +++ b/3250/CH6/EX6.3/Ex6_3.txt @@ -0,0 +1,3 @@ + # PROBLEM 6.3 #
+
+ Current required = 2240 amp
\ No newline at end of file diff --git a/3250/CH6/EX6.4/Ex6_4.sce b/3250/CH6/EX6.4/Ex6_4.sce new file mode 100755 index 000000000..697eea215 --- /dev/null +++ b/3250/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,40 @@ +clc +// Given that +I = 1000 // Current in amp +p1 = 72.5 // Percentage(by weight) of Ni in Nimonic 75 alloy +p2 = 19.5 // Percentage(by weight) of Cr in Nimonic 75 alloy +p3 = 5 // Percentage(by weight) of Fe in Nimonic 75 alloy +p4 = 0.4 // Percentage(by weight) of Ti in Nimonic 75 alloy +p5 = 1 // Percentage(by weight) of Si in Nimonic 75 alloy +p6 = 1 // Percentage(by weight) of Mn in Nimonic 75 alloy +p7 = 06 // Percentage(by weight) of Cu in Nimonic 75 alloy +// Sample Problem 4 on page no. 345 +printf("\n # PROBLEM 6.4 # \n") +// From the table 6.3 given in the book +D1 = 8.9 // Density of Ni in g/cm^3 +D2 = 7.19 // Density of Cr in g/cm^3 +D3 = 7.86 // Density of Fe in g/cm^3 +D4 = 4.51 // Density of Ti in g/cm^3 +D5 = 2.33 // Density of Si in g/cm^3 +D6 = 7.43 // Density of Mn in g/cm^3 +D7 = 8.96 // Density of Cu in g/cm^3 +A1 = 58.71 // Gram atomic weight of Ni in gm +A2 = 51.99 // Gram atomic weight of Cr in gm +A3 = 55.85 // Gram atomic weight of Fe in gm +A4 = 47.9 // Gram atomic weight of Ti in gm +A5 = 28.09 // Gram atomic weight of Si in gm +A6 = 54.94 // Gram atomic weight of Mn in gm +A7 = 63.57 // Gram atomic weight of Cu in gm +Z1 = 2 // Valence of dessolation for Ni +Z2 = 2 // Valence of dessolation for Cr +Z3 = 2 // Valence of dessolation for Fe +Z4 = 3 // Valence of dessolation for Ti +Z5 = 4 // Valence of dessolation for Si +Z6 = 2 // Valence of dessolation for Mn +Z7 = 1 // Valence of dessolation for Cu +// Above values are given in table 6.3 in the book +D = 100/((p1/D1)+(p2/D2)+(p3/D3)+(p4/D4)+(p5/D5)+(p6/D6)+(p7/D7)) +Q = ((0.1035*(10^-2))/D)*(1/((p1*Z1/A1)+(p2*Z2/A2)+(p3*Z3/A3)+(p4*Z4/A4)+(p5*Z5/A5)+(p6*Z6/A6)+(p7*Z7/A7))) +R = Q*I*60 +printf("\n Removal rate = %f cm^3/min",R) + diff --git a/3250/CH6/EX6.4/Ex6_4.txt b/3250/CH6/EX6.4/Ex6_4.txt new file mode 100755 index 000000000..820a17025 --- /dev/null +++ b/3250/CH6/EX6.4/Ex6_4.txt @@ -0,0 +1,3 @@ + # PROBLEM 6.4 #
+
+ Removal rate = 2.152707 cm^3/min
\ No newline at end of file diff --git a/3250/CH6/EX6.5/Ex6_5.sce b/3250/CH6/EX6.5/Ex6_5.sce new file mode 100755 index 000000000..81cd16fe8 --- /dev/null +++ b/3250/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,16 @@ +clc +// Given that +V = 10 // DC supply voltage in Volt +k = 0.2 // Conductivity of electrolyte in ohm^-1-cm^-1 +f = 0.1 // Feed rate in m/min +Vo = 1.5 // Total overvoltage in Volt +F = 96500 // Faraday constant in coulombs per mole +// Sample Problem 5 on page no. 352 +printf("\n # PROBLEM 6.5 # \n") +A = 55.85 // Atomic gram weight of iron in gm +Z = 2 // Valency of dissolation of iron +rho = 7.86 // Density of iron in gm/cm^3 +Yc = k*A*(V-Vo)/(rho*Z*F*(f/60)) +printf("\n Equilibrium gap = %f cm",Yc) + +// Answer in the book is given as 0.04 cm diff --git a/3250/CH6/EX6.5/Ex6_5.txt b/3250/CH6/EX6.5/Ex6_5.txt new file mode 100755 index 000000000..5eb7ed8fc --- /dev/null +++ b/3250/CH6/EX6.5/Ex6_5.txt @@ -0,0 +1,3 @@ + # PROBLEM 6.5 #
+
+ Equilibrium gap = 0.037553 cm
\ No newline at end of file diff --git a/3250/CH6/EX6.6/Ex6_6.sce b/3250/CH6/EX6.6/Ex6_6.sce new file mode 100755 index 000000000..f2d86b1f9 --- /dev/null +++ b/3250/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,17 @@ +clc +// Given that +S_I1 = 5 // Surface irregulation in micro meter +S_I2 = 8 // Surface irregulation in micro meter +V = 12 // DC supply voltage in Volt +k = 0.2 // Conductivity of electrolyte in ohm^-1-cm^-1 +Vo = 1.5 // Total overvoltage in Volt +F = 96500 // Faraday constant in coulombs per mole +// Sample Problem 6 on page no. 353 +printf("\n # PROBLEM 6.6 # \n") +Y_min = (S_I1+S_I2)*(10^(-4)) +A = 55.85 // Atomic gram weight of iron in gm +Z = 2 // Valency of dissolation of iron +D = 7.86 // Density of iron in gm/cm^3 +f_max = (k*A*(V-Vo)/(Z*D*F*Y_min))*60 +printf("\n Largest passible feed rate = %f mm/min",f_max*10) + diff --git a/3250/CH6/EX6.6/Ex6_6.txt b/3250/CH6/EX6.6/Ex6_6.txt new file mode 100755 index 000000000..31798e145 --- /dev/null +++ b/3250/CH6/EX6.6/Ex6_6.txt @@ -0,0 +1,3 @@ + # PROBLEM 6.6 #
+
+ Largest passible feed rate = 35.683752 mm/min
\ No newline at end of file diff --git a/3250/CH6/EX6.7/Ex6_7.sce b/3250/CH6/EX6.7/Ex6_7.sce new file mode 100755 index 000000000..5f600821a --- /dev/null +++ b/3250/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,29 @@ +clc +// Given that +f = 0.2 // Feed rate in cm/min +l = 2.54 // Length of tool face in cm +w = 2.54 // Width of tool face in cm +T_b = 95 // Boiling temperature of electrolyte in °C +Nita = 0.876e-3 // Viscosity of electrolyte in kg/m-sec +D_e = 1.088 // Density of electrolyte in g/cm^3 +c = .997 // Specific heat of electrolyte +V = 10 // DC supply voltage in Volt +k = 0.2 // Conductivity of electrolyte in ohm^-1-cm^-1 +T = 35 // Ambient temperature in °C +Vo = 1.5 // Total overvoltage in Volt +F = 96500 // Faraday constant in coulombs per mole +// Sample Problem 7 on page no. 355 +printf("\n # PROBLEM 6.7 # \n") +A = 55.85 // Atomic gram weight of iron in gm +Z = 2 // Valency of dissolation of iron +D = 7.86 // Density of iron in gm/cm^3 +Ye = k*A*(V-Vo)*60/(D*Z*F*f) +J = k*(V-Vo)/(Ye) +D_T = T_b -T +v = (J^2)*(l)/(k*D_T*D_e*c) +Re = ((D_e*v*2*Ye)/Nita)*(0.1) +p = 0.3164*D_e*(v^2)*l/(4*Ye*(Re^0.25))*(10^-4) +A = l*w +F = p*A*(10^-1)*(1/2) +printf("\n Total force acting on the tool = %d N",F) +// Answer in the book is given as 79 N diff --git a/3250/CH6/EX6.7/Ex6_7.txt b/3250/CH6/EX6.7/Ex6_7.txt new file mode 100755 index 000000000..214bb5628 --- /dev/null +++ b/3250/CH6/EX6.7/Ex6_7.txt @@ -0,0 +1,3 @@ + # PROBLEM 6.7 #
+
+ Total force acting on the tool = 103 N
\ No newline at end of file diff --git a/3250/CH6/EX6.8/Ex6_8.sce b/3250/CH6/EX6.8/Ex6_8.sce new file mode 100755 index 000000000..2e206f6d5 --- /dev/null +++ b/3250/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,19 @@ +clc +x = poly(0,"x") +// Given that +y = 10+0.3*x-0.05*x^2//Equation of geometry of workpiece surface +V = 15 // Applied potential in Volt +f = 0.75 // Feed velocity in cm/min +k = 0.2 // Conductivity of electrolyte in ohm^-1-cm^-1 +Vo = 0.67 // Total overvoltage in Volt +F = 96500 // Faraday constant in coulombs per mole +// Sample Problem 8 on page no. 361 +printf("\n # PROBLEM 6.8 # \n") +A = 63.57 // Atomic gram weight of copper in gm +Z = 1 // Valency of dissolation of copper +D = 8.96 // Density of copper in gm/cm^3 +lambda = k*A*(V-Vo)/(D*Z*F) +r = lambda/(f/(10*60)) +Y = 10 + 0.3*(x-(r*((0.3-0.1*x)/(1-0.1*r)))) - 0.05*(x-(r*((0.3-.1*x)/(1-0.1*r))))^2 - r +printf("\n The equation of required tool geometry is :-\n y =") +disp(Y) diff --git a/3250/CH6/EX6.8/Ex6_8.txt b/3250/CH6/EX6.8/Ex6_8.txt new file mode 100755 index 000000000..983c4a866 --- /dev/null +++ b/3250/CH6/EX6.8/Ex6_8.txt @@ -0,0 +1,8 @@ +
+ # PROBLEM 6.8 #
+
+ The equation of required tool geometry is :-
+ y =
+ 2
+ 9.8158651 + 0.3103759x - 0.0517293x
+
\ No newline at end of file diff --git a/3250/CH6/EX6.9/Ex6_9.sce b/3250/CH6/EX6.9/Ex6_9.sce new file mode 100755 index 000000000..dd148f607 --- /dev/null +++ b/3250/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,18 @@ +clc +// Given that +a = 10 // Side length of a square hole in mm +t = 5 // Thickness of low carbon steel plate in mm +R = 50 // Resistance in relaxation circuit in ohm +C = 10 // Capacitance in relaxation circuit in micro F +V = 200 // Supply voltage in Volt +V_ = 150 // Minimum required voltage for discharge in Volt +// Sample Problem 9 on page no. 378 +printf("\n # PROBLEM 6.9 # \n") +E = (1/2)*C*(10^-6)*(V_^2) +tc = R*C*(10^-6)*log(V/(V-V_)) +W = (E/tc)*(10^-3) +v = t*a^2 +Q = 27.4*(W^(1.54)) +T = v/Q +printf("\n The time required to complete the drilling operation = %d min",T) +// Answer in the book is given as 306 min diff --git a/3250/CH6/EX6.9/Ex6_9.txt b/3250/CH6/EX6.9/Ex6_9.txt new file mode 100755 index 000000000..78576f43d --- /dev/null +++ b/3250/CH6/EX6.9/Ex6_9.txt @@ -0,0 +1,3 @@ + # PROBLEM 6.9 #
+
+ The time required to complete the drilling operation = 300 min
\ No newline at end of file diff --git a/3250/CH7/EX7.1/Ex7_1.sce b/3250/CH7/EX7.1/Ex7_1.sce new file mode 100755 index 000000000..7c0006ab6 --- /dev/null +++ b/3250/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,10 @@ +clc +// Given that +F = 4e-6 // Maximum feature dimension in meter +t = 5e-6 // Photorist thickness in meter +g = 25e-6 // Allowable gap between the mask and the resist meter +// Sample Problem 1 on page no. 432 +printf("\n # PROBLEM 7.1 # \n") +lambda = (F^2)/(t+g) +printf("\n Maximum allowable wavelength of the exposing light = %d nm",lambda*(10^9)) + diff --git a/3250/CH7/EX7.1/Ex7_1.txt b/3250/CH7/EX7.1/Ex7_1.txt new file mode 100755 index 000000000..b1871f35e --- /dev/null +++ b/3250/CH7/EX7.1/Ex7_1.txt @@ -0,0 +1,4 @@ +
+ # PROBLEM 7.1 #
+
+ Maximum allowable wavelength of the exposing light = 533 nm
\ No newline at end of file diff --git a/3250/CH7/EX7.2/Ex7_2.sce b/3250/CH7/EX7.2/Ex7_2.sce new file mode 100755 index 000000000..d27f7efc5 --- /dev/null +++ b/3250/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,9 @@ +clc +// Given that +d = 5 // Diameter of hole in micro meter +h = 100 // Depth of hole in micro meter +// Sample Problem 2 on page no. 440 +printf("\n # PROBLEM 7.2 # \n") +t = 31.58*(d*(exp(h/(60*d))-1)) +printf("\n Time required to machine the hole = %f min",t) + diff --git a/3250/CH7/EX7.2/Ex7_2.txt b/3250/CH7/EX7.2/Ex7_2.txt new file mode 100755 index 000000000..e38e73b9a --- /dev/null +++ b/3250/CH7/EX7.2/Ex7_2.txt @@ -0,0 +1,3 @@ + # PROBLEM 7.2 #
+
+ Time required to machine the hole = 62.467202 min
\ No newline at end of file diff --git a/3250/CH7/EX7.3/Ex7_3.sce b/3250/CH7/EX7.3/Ex7_3.sce new file mode 100755 index 000000000..68c9241d4 --- /dev/null +++ b/3250/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,9 @@ +clc +// Given that +J = 2 // The threshold value of dose in kJ/cm^3 +h = 300 // Height in micro meter +// Sample Problem 3 on page no. 448 +printf("\n # PROBLEM 7.3 # \n") +J_o = J*(exp(0.1*sqrt(h))) +printf("\n The minimum level of exposure of the PMMA surface = %f kJ/cm^3",J_o) + diff --git a/3250/CH7/EX7.3/Ex7_3.txt b/3250/CH7/EX7.3/Ex7_3.txt new file mode 100755 index 000000000..092111766 --- /dev/null +++ b/3250/CH7/EX7.3/Ex7_3.txt @@ -0,0 +1,3 @@ + # PROBLEM 7.3 #
+
+ The minimum level of exposure of the PMMA surface = 11.304467 kJ/cm^3
\ No newline at end of file diff --git a/3250/CH7/EX7.4/Ex7_4.sce b/3250/CH7/EX7.4/Ex7_4.sce new file mode 100755 index 000000000..c9c7f2b9c --- /dev/null +++ b/3250/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,12 @@ +clc +// Given that +J_ = 2 // The threshold value of dose in kJ/cm^3 +J = 15 // The dose of top surface in kJ/cm^3 +x_ = 300 // Depth below the surface in micro meter +// Sample Problem 4 on page no. 4 +printf("\n # PROBLEM 7.4 # \n") +function y=f(x),y = 3/((J*(exp(-0.1*sqrt(x))))^(1.6)-3), +endfunction +t = intg(0,x_,f) +printf("\n The time required to develop the PMMA resist = %d min",t) + diff --git a/3250/CH7/EX7.4/Ex7_4.txt b/3250/CH7/EX7.4/Ex7_4.txt new file mode 100755 index 000000000..4fd4dd4d2 --- /dev/null +++ b/3250/CH7/EX7.4/Ex7_4.txt @@ -0,0 +1,3 @@ + # PROBLEM 7.4 #
+
+ The time required to develop the PMMA resist = 157 min
\ No newline at end of file |