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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /3308 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '3308')
44 files changed, 725 insertions, 0 deletions
diff --git a/3308/CH10/EX10.1/Ex10_1.sce b/3308/CH10/EX10.1/Ex10_1.sce new file mode 100755 index 000000000..dcc72e4c6 --- /dev/null +++ b/3308/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,30 @@ +clc
+// Given that
+//three metal piece being cast have the same volume but different shapes
+//shapes are sphere,cube,cylinder(height=diameter)
+
+// Sample Problem on page no. 252
+
+printf("\n #solidification time for various shapes# \n")
+
+//solidification time is inversely proportional to the square of surface area
+
+//for sphere
+r=(3/(4*3.14))^(1/3)//radius of the sphere from volume of sphere v=(4*3.14*r^3)/3
+A=4*3.14*((r)^2)
+time1=1/(A)^2
+printf("\n the solidification time for the sphere is %fC",time1)
+
+//for cube
+a=1//edge of the cube
+A=6*a^2
+time2=1/(A)^2
+printf("\n the solidification time for the cube is %fC",time2)
+
+//for cylinder
+//given height =diameter
+//radius=2*height
+r=(1/(2*3.14))^(1/3)//radius of the cylinder from volume of the cylinder v=3.14*r^2*h
+A=(6*3.14*(r^2)) //area of the cylinder = (2*3.14*radius^2) + (2*3.14*radius*height)
+time3=1/(A)^2
+printf("\n the solidification time for the sphere is %fC",time3)
diff --git a/3308/CH10/EX10.1/Ex10_1.txt b/3308/CH10/EX10.1/Ex10_1.txt new file mode 100755 index 000000000..edb84e7de --- /dev/null +++ b/3308/CH10/EX10.1/Ex10_1.txt @@ -0,0 +1,6 @@ +
+ #solidification time for various shapes#
+
+ the solidification time for the sphere is 0.042774C
+ the solidification time for the cube is 0.027778C
+ the solidification time for the sphere is 0.032643C
\ No newline at end of file diff --git a/3308/CH13/EX13.1/Ex13_1.sce b/3308/CH13/EX13.1/Ex13_1.sce new file mode 100755 index 000000000..e4b2905c9 --- /dev/null +++ b/3308/CH13/EX13.1/Ex13_1.sce @@ -0,0 +1,33 @@ +clc
+// Given that
+w=9 //in inch width of thee strip
+ho=1 //in inch initial thickness of the strip
+hf=0.80 //in inch thickness of the strip after one pass
+r=12 //in inch roll radius
+N=100 //in rpm
+
+// Sample Problem on page no. 323
+
+printf("\n #Calculation of roll force and torque# \n")
+
+L=(r*(ho-hf))^(1/2)
+
+E=log(1/hf)//absolute value of true strain
+
+Y=26000 //in psi average stress from the data in the book
+F=L*w*Y // roll force
+F1=F*4.448/(10^6)//in mega newton
+printf("\n\nRoll force = %f MN ",F1)
+
+//answer in the book is round off and given 363000lb
+
+P=(2*3.14*F*L*N)/(33000*12)
+P1=P*7.457*(10^2)/(10^3)//in KW
+printf("\n\npower per roll = %f KW ",P1)
+
+//answer in the book is 670 KW due to round off of the roll force
+
+Tp=2*P1//total power
+printf("\n\nTotal power = %f KW ",Tp)
+
+//answer in the book is 1340KW due to round off of the roll force
diff --git a/3308/CH13/EX13.1/Ex13_1.txt b/3308/CH13/EX13.1/Ex13_1.txt new file mode 100755 index 000000000..a6db424ec --- /dev/null +++ b/3308/CH13/EX13.1/Ex13_1.txt @@ -0,0 +1,8 @@ +#Calculation of roll force and torque#
+
+
+Roll force = 1.612450 MN
+
+power per roll = 664.133978 KW
+
+Total power = 1328.267956 KW
\ No newline at end of file diff --git a/3308/CH14/EX14.1/Ex14_1.sce b/3308/CH14/EX14.1/Ex14_1.sce new file mode 100755 index 000000000..44bdaf348 --- /dev/null +++ b/3308/CH14/EX14.1/Ex14_1.sce @@ -0,0 +1,22 @@ +clc
+// Given that
+d=150//in mm Diameter of the solid cylinder
+Hi=100 //in mm Height of the cylinder
+u=0.2 // Cofficient of friction
+
+// Sample Problem on page no. 344
+
+printf("\n # Calculation of forging force # \n")
+
+//cylinder is reduced in height by 50%
+Hf=100/2
+//Volume before deformation= Volume after deformation
+r=sqrt((3.14*75^2*100)/(3.14*50))//r is the final radius of the cylinder
+E=log(Hi/Hf)//absolute value of true strain
+//given that cylinder is made of 304 stainless steel
+Yf=1000 //in Mpa flow stress of the material from data in the book
+F = Yf*(10^6)*3.14*(r^2)*10^-6*(1+((2*u*r)/(3*Hf)))//Forging Force
+F1=F/(10^6)
+printf("\n\n Forging force = %d MN",F1)
+
+
diff --git a/3308/CH14/EX14.1/Ex14_1.txt b/3308/CH14/EX14.1/Ex14_1.txt new file mode 100755 index 000000000..56865068e --- /dev/null +++ b/3308/CH14/EX14.1/Ex14_1.txt @@ -0,0 +1,4 @@ + # Calculation of forging force #
+
+
+ Forging force = 45 MN
\ No newline at end of file diff --git a/3308/CH15/EX15.1/Ex15_1.sce b/3308/CH15/EX15.1/Ex15_1.sce new file mode 100755 index 000000000..13264633e --- /dev/null +++ b/3308/CH15/EX15.1/Ex15_1.sce @@ -0,0 +1,16 @@ +clc
+// Given that
+di=5//in inch Diameter of the round billet
+df=2//in inch Diameter of the round billet after extrusion
+
+// Sample Problem on page no. 372
+
+printf("\n # Calculation of force in Hot Extrusion# \n")
+
+//As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book
+k=35000//in psi
+F=3.14*(di/2)^2*k*log((3.14*(di^2))/(3.14*(df^2)))
+F1=F*4.448/(10^6)
+printf("\n\n Extrusion force=%f MN",F1)
+
+//Answer in the book is approximated to 5.5MN
diff --git a/3308/CH15/EX15.1/Ex15_1.txt b/3308/CH15/EX15.1/Ex15_1.txt new file mode 100755 index 000000000..4fda21cac --- /dev/null +++ b/3308/CH15/EX15.1/Ex15_1.txt @@ -0,0 +1,5 @@ +
+ # Calculation of force in Hot Extrusion#
+
+
+ Extrusion force=5.598940 MN
\ No newline at end of file diff --git a/3308/CH16/EX16.1/Ex16_1.sce b/3308/CH16/EX16.1/Ex16_1.sce new file mode 100755 index 000000000..a9e511c5b --- /dev/null +++ b/3308/CH16/EX16.1/Ex16_1.sce @@ -0,0 +1,16 @@ +clc
+// Given that
+d=1//in inch Diameter of the hole
+T=(1/8)//in inch thickness of the sheet
+
+// Sample Problem on page no. 396
+
+printf("\n # Calculation of Punch Force# \n")
+
+UTS=140000//in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V
+L=3.14*d//total length sheared which is the perimeter of the hole
+F=0.7*T*L*UTS
+F1=F*4.448/(10^6)
+printf("\n\n Extrusion force=%f MN",F1)
+
+//Answer in the book is approximated to 0.17MN
diff --git a/3308/CH16/EX16.1/Ex16_1.txt b/3308/CH16/EX16.1/Ex16_1.txt new file mode 100755 index 000000000..73978664a --- /dev/null +++ b/3308/CH16/EX16.1/Ex16_1.txt @@ -0,0 +1,4 @@ + # Calculation of Punch Force#
+
+
+ Extrusion force=0.171092 MN
\ No newline at end of file diff --git a/3308/CH17/EX17.1/Ex17_1.sce b/3308/CH17/EX17.1/Ex17_1.sce new file mode 100755 index 000000000..65463777b --- /dev/null +++ b/3308/CH17/EX17.1/Ex17_1.sce @@ -0,0 +1,27 @@ +clc
+// Given that
+L=20//in mm Final length of the ceramic part
+//Linear shrinkage during drying and firing is 7% and 6% respectively
+Sd=0.070//Linear shrinkage during drying
+Sf=0.06//Linear shrinkage during firing
+
+// Sample Problem on page no. 466
+
+printf("\n # Dimensional changes during the shaping of ceramic components # \n")
+
+//part (a)
+
+Ld=L/(1-Sf)//dried length
+Lo=(1+Sd)*Ld//initial length
+printf("\n\n Initial Length=%f mm",Lo)
+
+//Answer in the book is approximated to 22.77mm
+
+//part(b)
+
+Pf=0.03//Fired Porosity
+r = (1-Pf)// Where r = Va/Vf
+R = 1/((1-Sf)^3)// Where R = Vd/Vf
+Pd = (1-r/R)
+printf("\n\nDried porosity is %d percent",Pd*100)
+
diff --git a/3308/CH17/EX17.1/Ex17_1.txt b/3308/CH17/EX17.1/Ex17_1.txt new file mode 100755 index 000000000..ba3af5257 --- /dev/null +++ b/3308/CH17/EX17.1/Ex17_1.txt @@ -0,0 +1,7 @@ +
+ # Dimensional changes during the shaping of ceramic components #
+
+
+ Initial Length=22.765957 mm
+
+Dried porosity is 19 percent
\ No newline at end of file diff --git a/3308/CH18/EX18.1/Ex18_1.sce b/3308/CH18/EX18.1/Ex18_1.sce new file mode 100755 index 000000000..06b725a81 --- /dev/null +++ b/3308/CH18/EX18.1/Ex18_1.sce @@ -0,0 +1,19 @@ +clc
+// Given that
+W=400//in mm Lateral(width) Dimension of a plastic shopping bag
+
+// Sample Problem on page no. 484
+
+printf("\n # Blown Film # \n")
+
+//part(a)
+
+P=2*W//in mm Perimeter of bag
+D=P/3.14//in mm blown diameter calculated from Permeter=3.14*diameter
+//Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5
+Dd=D/2.5//Extrusion die diameter
+printf("\n\n Extrusion Die Diameter =%d mm",Dd)
+
+//Answer in the book is approximated to 100mm
+
+//part(b) is theoritical
diff --git a/3308/CH18/EX18.1/Ex18_1.txt b/3308/CH18/EX18.1/Ex18_1.txt new file mode 100755 index 000000000..2d442624b --- /dev/null +++ b/3308/CH18/EX18.1/Ex18_1.txt @@ -0,0 +1,4 @@ + # Blown Film #
+
+
+ Extrusion Die Diameter =101 mm
\ No newline at end of file diff --git a/3308/CH18/EX18.2/Ex18_2.sce b/3308/CH18/EX18.2/Ex18_2.sce new file mode 100755 index 000000000..7b14b1ffd --- /dev/null +++ b/3308/CH18/EX18.2/Ex18_2.sce @@ -0,0 +1,19 @@ +clc
+// Given that
+W=250//in ton Weight of injection moulding machine
+d=4.5//in inch diameter of spur gear
+t=0.5//in inch thickness of spur gear
+//Gears have a fine tooth profile
+
+// Sample Problem on page no. 488
+
+printf("\n # Injection Molding of Parts # \n")
+
+//because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi
+p=15//inKsi
+A=(3.14*(d^2))/4//in inch^2 area of the gear
+F=A*15*1000
+n=(W*2000)/F //weight is converted into lb by multiplying it by 2000
+printf("\n\n Number of gears that can be injected =%d",n)
+
+// Second part of this question is theoritical
diff --git a/3308/CH18/EX18.2/Ex18_2.txt b/3308/CH18/EX18.2/Ex18_2.txt new file mode 100755 index 000000000..b473416ea --- /dev/null +++ b/3308/CH18/EX18.2/Ex18_2.txt @@ -0,0 +1,4 @@ + # Injection Molding of Parts #
+
+
+ Number of gears that can be injected =2
\ No newline at end of file diff --git a/3308/CH2/EX2.1/Ex2_1.sce b/3308/CH2/EX2.1/Ex2_1.sce new file mode 100755 index 000000000..fa06cfb68 --- /dev/null +++ b/3308/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,24 @@ +
+clc
+// Given that
+//True stress=100000*(True strain)^0.5
+
+// Sample Problem on page no. 63
+
+printf("\n # Calculation of Ultimate Tensile Strength # \n")
+//from the data given
+n=0.5
+E=0.5
+K=100000
+Truestress=K*((E)^n)
+//let An(area of neck)/Ao=t
+//from log(Ao/An)=n
+t=exp(-n)
+UTS=Truestress*exp(-n)//from the expression UTS= P/Ao where P(Maximum Load)=Truestress*An
+printf("\n\n Ultimate Tensile Strength = %f psi",UTS)
+//answer in the book is approximated to 42850 psi
+
+
+
+
+
diff --git a/3308/CH2/EX2.1/Ex2_1.txt b/3308/CH2/EX2.1/Ex2_1.txt new file mode 100755 index 000000000..895279878 --- /dev/null +++ b/3308/CH2/EX2.1/Ex2_1.txt @@ -0,0 +1,4 @@ + # Calculation of Ultimate Tensile Strength #
+
+
+ Ultimate Tensile Strength = 42888.194248 psi
\ No newline at end of file diff --git a/3308/CH20/EX20.1/Ex20_1.sce b/3308/CH20/EX20.1/Ex20_1.sce new file mode 100755 index 000000000..93bd6b3d2 --- /dev/null +++ b/3308/CH20/EX20.1/Ex20_1.sce @@ -0,0 +1,22 @@ +clc
+// Given that
+to=0.005//in inch depth of cut
+V=400//in ft/min cutting speed
+X=10//in degree rake angle
+w=0.25//in inch width of cut
+tc=0.009//in inch chip thickness
+Fc=125//in lb Cutting force
+Ft=50//in lb thrust force
+
+// Sample Problem on page no. 548
+
+printf("\n # Relative Energies in cutting # \n")
+
+r=to/tc//cutting ratio
+R=sqrt((Ft^2)+(Fc^2))
+B=acosd(Fc/R)+X//friction angle
+F=R*sind(B)
+P=((F*r)/Fc)*100
+printf("\n\n Percentage of total energy going into overcoming friction =%d pecrent",P)
+
+//Answer in the book is approximated to 32 due to approximation in calculation of R and B
diff --git a/3308/CH20/EX20.1/Ex20_1.txt b/3308/CH20/EX20.1/Ex20_1.txt new file mode 100755 index 000000000..7c219531f --- /dev/null +++ b/3308/CH20/EX20.1/Ex20_1.txt @@ -0,0 +1,5 @@ +
+ # Relative Energies in cutting #
+
+
+ Percentage of total energy going into overcoming friction =31 pecrent
\ No newline at end of file diff --git a/3308/CH20/EX20.2/Ex20_2.sce b/3308/CH20/EX20.2/Ex20_2.sce new file mode 100755 index 000000000..d93338bc0 --- /dev/null +++ b/3308/CH20/EX20.2/Ex20_2.sce @@ -0,0 +1,17 @@ +clc
+// Given that
+n=0.5//exponent that depends on tool and workpiece material
+C=400//constant
+
+// Sample Problem on page no. 555
+
+printf("\n # Increasing tool life by Reducing the Cutting Speed # \n")
+
+V1=poly(0,"V1")
+r=0.5// it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool
+//let t=T2/T1 where T1 and T2 are the initial and final tool life
+t=1/(r^(1/n))//from the relation V1*(T1^n)=V2*(T2^n)
+P=(t-1)*100
+printf("\n\n Percent increase in tool life =%d Percent",P)
+
+
diff --git a/3308/CH20/EX20.2/Ex20_2.txt b/3308/CH20/EX20.2/Ex20_2.txt new file mode 100755 index 000000000..6e441f1e6 --- /dev/null +++ b/3308/CH20/EX20.2/Ex20_2.txt @@ -0,0 +1,4 @@ + # Increasing tool life by Reducing the Cutting Speed #
+
+
+ Percent increase in tool life =300 Percent
\ No newline at end of file diff --git a/3308/CH22/EX22.1/Ex22_1.sce b/3308/CH22/EX22.1/Ex22_1.sce new file mode 100755 index 000000000..4f24bf3be --- /dev/null +++ b/3308/CH22/EX22.1/Ex22_1.sce @@ -0,0 +1,36 @@ +clc
+// Given that
+l=6//in inch Length of rod
+di=1/2//in inch initial diameter of rod
+df=0.480//in inch final diameter of rod
+N=400//in rpm spindle rotation
+Vt=8//in inch/minute axial speed of the tool
+
+// Sample Problem on page no. 600
+
+printf("\n # Material Removal Rate and Cutting Force in Turning # \n")
+
+V=3.14*di*N
+printf("\n\n Cutting speed=%d in/min",V)
+
+v1=3.14*df*N//cutting speed from machined diameter
+d=(di-df)/2//depth of cut
+f=Vt/N//feed
+Davg=(di+df)/2
+MRR=3.14*Davg*d*f*N
+printf("\n\n Material Removal Rate %f=in^3/min",MRR)
+
+t=l/(f*N)
+printf("\n\n Cutting time=%f min",t)
+
+P=(4/2.73)*MRR//average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3
+printf("\n\n Cutting power=%f hp",P)
+
+Fc=((P*396000)/(N*2*3.14))/(Davg/2)
+printf("\n\n Cutting force=%d lb",Fc)
+
+//answer in the book is given 118 lb due to approximation
+
+
+
+
diff --git a/3308/CH22/EX22.1/Ex22_1.txt b/3308/CH22/EX22.1/Ex22_1.txt new file mode 100755 index 000000000..9b0ca6b2e --- /dev/null +++ b/3308/CH22/EX22.1/Ex22_1.txt @@ -0,0 +1,12 @@ + # Material Removal Rate and Cutting Force in Turning #
+
+
+ Cutting speed=628 in/min
+
+ Material Removal Rate 0.123088=in^3/min
+
+ Cutting time=0.750000 min
+
+ Cutting power=0.180349 hp
+
+ Cutting force=116 lb
\ No newline at end of file diff --git a/3308/CH22/EX22.2/Ex22_2.sce b/3308/CH22/EX22.2/Ex22_2.sce new file mode 100755 index 000000000..d2794ff75 --- /dev/null +++ b/3308/CH22/EX22.2/Ex22_2.sce @@ -0,0 +1,20 @@ +clc
+// Given that
+d=10//in mm diameter of drill bit
+f=0.2//in mm/rev feed
+N=800//in rpm spindle rotation
+
+// Sample Problem on page no. 632
+
+printf("\n # Material Removal Rate and Torque in Drilling # \n")
+
+MRR=[((3.14*(d^2))/4)*f*N]/60
+printf("\n\n Material Removal Rate %d=mm^3/sec",MRR)
+
+//Answer in the book is given 210 mm^3/sec
+
+//from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken
+T=(MRR*0.5)/((N*2*3.14)/60)
+printf("\n\n Torque on the drill %f=Nm",T)
+
+
diff --git a/3308/CH22/EX22.2/Ex22_2.txt b/3308/CH22/EX22.2/Ex22_2.txt new file mode 100755 index 000000000..c0063062b --- /dev/null +++ b/3308/CH22/EX22.2/Ex22_2.txt @@ -0,0 +1,7 @@ +
+ # Material Removal Rate and Torque in Drilling #
+
+
+ Material Removal Rate 209=mm^3/sec
+
+ Torque on the drill 1.250000=Nm
\ No newline at end of file diff --git a/3308/CH23/EX23.1/Ex23_1.sce b/3308/CH23/EX23.1/Ex23_1.sce new file mode 100755 index 000000000..bf388a0b8 --- /dev/null +++ b/3308/CH23/EX23.1/Ex23_1.sce @@ -0,0 +1,34 @@ +clc
+// Given that
+l=12//in inch Length of block
+w=4//in inch width
+f=0.01//in inch/tooth feed
+d=1/8//in inch depth of cut
+D=2//in inch diameter of cutter
+n=20//no. of teeth
+N=100//in rpm spindle rotation
+Vt=8//in inch/minute axial speed of the tool
+
+// Sample Problem on page no. 600
+
+printf("\n # Material Removal Rate , Power required and Cutting Time in slab milling # \n")
+
+v=f*N*n
+MRR=w*d*v
+printf("\n\n Material Removal Rate = %d in^3/min",MRR)
+
+//for annealed mild steel unit power is taken as 1.1 hp min/in3
+P=1.1*MRR
+printf("\n\n Cutting power=%d hp",P)
+
+T=P*33000/(N*2*3.14)
+printf("\n\n Cutting torque=%d lb-ft",T)
+
+lc=sqrt(d*D)
+t=((l+lc)/20)*60
+printf("\n\n Cutting time=%f sec",t)
+
+
+
+
+
diff --git a/3308/CH23/EX23.1/Ex23_1.txt b/3308/CH23/EX23.1/Ex23_1.txt new file mode 100755 index 000000000..817ce4e24 --- /dev/null +++ b/3308/CH23/EX23.1/Ex23_1.txt @@ -0,0 +1,11 @@ +
+ # Material Removal Rate , Power required and Cutting Time in slab milling #
+
+
+ Material Removal Rate = 10 in^3/min
+
+ Cutting power=11 hp
+
+ Cutting torque=578 lb-ft
+
+ Cutting time=37.500000 sec
\ No newline at end of file diff --git a/3308/CH23/EX23.2/Ex23_2.sce b/3308/CH23/EX23.2/Ex23_2.sce new file mode 100755 index 000000000..40ca4ed31 --- /dev/null +++ b/3308/CH23/EX23.2/Ex23_2.sce @@ -0,0 +1,34 @@ +clc
+// Given that
+l=500//in mm Length
+w=60//in mm width
+v=0.6//in m/min
+d=3//in mm depth of cut
+D=150//in mm diameter of cutter
+n=10//no. of inserts
+N=100//in rpm spindle rotation
+
+// Sample Problem on page no. 655
+
+printf("\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n")
+
+MRR=w*d*v*1000
+printf("\n\n Material Removal Rate = %d mm3/min",MRR)
+
+lc=D/2
+t=((l+(2*lc))/((v*1000)/60)) // velocity is converted into mm/sec
+t1=t/60
+printf("\n\n Cutting time= %ff min",t1)
+
+f=(v*1000*60)/(60*N*n) // N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec
+printf("\n\n Feed per Tooth= %f mm/tooth",f)
+
+//for high strength aluminium alloy unit power is taken as 1.1 W s/mm3
+P=(1.1*MRR)/60 // MRR is converted into mm3/sec by dividing by 60
+P1=P/(1000)//in KW
+printf("\n\n Cutting power=%f KW",P1)
+
+
+
+
+
diff --git a/3308/CH23/EX23.2/Ex23_2.txt b/3308/CH23/EX23.2/Ex23_2.txt new file mode 100755 index 000000000..9174c0253 --- /dev/null +++ b/3308/CH23/EX23.2/Ex23_2.txt @@ -0,0 +1,11 @@ +
+ # Material Removal Rate , Power Required and Cutting Time in Face Milling #
+
+
+ Material Removal Rate = 108000 mm3/min
+
+ Cutting time= 1.083333f min
+
+ Feed per Tooth= 0.600000 mm/tooth
+
+ Cutting power=1.980000 KW
\ No newline at end of file diff --git a/3308/CH25/EX25.1/Ex25_1.sce b/3308/CH25/EX25.1/Ex25_1.sce new file mode 100755 index 000000000..a58514c32 --- /dev/null +++ b/3308/CH25/EX25.1/Ex25_1.sce @@ -0,0 +1,28 @@ +clc
+// Given that
+D=200//in mm Grinding Wheel diameter
+d=0.05//in mm depth of cut
+v=30//m/min workpiece velocity
+V=1800//in m/min wheel velocity
+
+// Sample Problem on page no. 713
+
+printf("\n # Chip Dimensions in Surface Grinding # \n")
+
+l=sqrt(D*d)
+l1=l/2.54*(10^-1)
+printf("\n\n Undeformed Chip Length = %f mm",l1)
+
+//the answer in the book is approximated to 0.13 in
+
+//assume
+C=2//in mm
+r=15
+t=sqrt(((4*v)/(V*C*r))*sqrt(d/D))
+t1=t/2.54*(10^-1)
+printf("\n\n Undeformed chip Thickness = %f in",t1)
+
+//the answer in the book is approximated to 0.00023in
+
+
+
diff --git a/3308/CH25/EX25.1/Ex25_1.txt b/3308/CH25/EX25.1/Ex25_1.txt new file mode 100755 index 000000000..982289851 --- /dev/null +++ b/3308/CH25/EX25.1/Ex25_1.txt @@ -0,0 +1,7 @@ +
+ # Chip Dimensions in Surface Grinding #
+
+
+ Undeformed Chip Length = 0.124499 mm
+
+ Undeformed chip Thickness = 0.000233 in
\ No newline at end of file diff --git a/3308/CH25/EX25.2/Ex25_2.sce b/3308/CH25/EX25.2/Ex25_2.sce new file mode 100755 index 000000000..d2168cb10 --- /dev/null +++ b/3308/CH25/EX25.2/Ex25_2.sce @@ -0,0 +1,32 @@ +clc
+// Given that
+D=10//in inch Grinding Wheel diameter
+N=4000//in rpm
+w=1//in inch
+d=0.002//in inch depth of cut
+v=60//inch/min feed rate of the workpiece
+
+// Sample Problem on page no. 715
+
+printf("\n # force in Surface Grinding # \n")
+
+Mrr=d*w*v//material removal rate
+//for low carbon steel , the specific energy is 15hp min/in3
+u=15//in hp min/in3
+P=u*Mrr*396000//in lb/min
+Fc = P/(2*3.14*N*(D/2))
+
+printf("\n\n Cutting Force = %f lb",Fc)
+// Answer in the book is approximated to 5.7 lb
+
+// from the experimental data in book thrust force is taken as 30% higher than cutting force
+Fn = Fc+(30/100)*Fc
+
+printf("\n\n Thrust Force = %f lb",Fn)
+// Answer in the book is approximated to 7.4 lb
+
+
+
+
+
+
diff --git a/3308/CH25/EX25.2/Ex25_2.txt b/3308/CH25/EX25.2/Ex25_2.txt new file mode 100755 index 000000000..1d57b5c6b --- /dev/null +++ b/3308/CH25/EX25.2/Ex25_2.txt @@ -0,0 +1,7 @@ +
+ # force in Surface Grinding #
+
+
+ Cutting Force = 5.675159 lb
+
+ Thrust Force = 7.377707 lb
\ No newline at end of file diff --git a/3308/CH28/EX28.1/Ex28_1.sce b/3308/CH28/EX28.1/Ex28_1.sce new file mode 100755 index 000000000..d6752b442 --- /dev/null +++ b/3308/CH28/EX28.1/Ex28_1.sce @@ -0,0 +1,34 @@ +clc
+// Given that
+t=1//in mm thickness of chip
+I=5000//in Ampere current
+T=0.1//in sec
+d=5//in mm diameter of electrode
+
+
+// Sample Problem on page no. 805
+
+printf("\n # Heat Generated in Spot Welding # \n")
+
+//It is assumed in the book that effective restiance = 200 micro ohm
+R=200*(10^-6)
+H=(I^2)*R*T
+
+printf("\n\n Heat Generated = %d J",H)
+
+// It is assumed in the book that
+V=30//in mm3 volume
+D=0.008//in g/mm3 density
+M=D*V
+//Heat required to melt 1 g of steel is about 1400J
+m1=1400*M
+printf("\n\n Heat Required to melt weld nugget = %d J",m1)
+
+m2=H-m1
+printf("\n\n Heat Dissipitated into the metal surrounding the nugget = %d J",m2)
+
+
+
+
+
+
diff --git a/3308/CH28/EX28.1/Ex28_1.txt b/3308/CH28/EX28.1/Ex28_1.txt new file mode 100755 index 000000000..7a2987c6c --- /dev/null +++ b/3308/CH28/EX28.1/Ex28_1.txt @@ -0,0 +1,9 @@ +
+ # Heat Generated in Spot Welding #
+
+
+ Heat Generated = 500 J
+
+ Heat Required to melt weld nugget = 336 J
+
+ Heat Dissipitated into the metal surrounding the nugget = 164 J
\ No newline at end of file diff --git a/3308/CH32/EX32.1/Ex32_1.sce b/3308/CH32/EX32.1/Ex32_1.sce new file mode 100755 index 000000000..cf732b00b --- /dev/null +++ b/3308/CH32/EX32.1/Ex32_1.sce @@ -0,0 +1,22 @@ +clc
+// Given that
+hi=10//in mm height of specimen
+ODi=30//in mm outside diameter
+IDi=15//in mm inside diameter
+ODf=38//in mm outside diameter after deformaton
+//Specimen is reduced in thickness by 50%
+hf=(50/100)*hi
+
+// Sample Problem on page no. 886
+
+printf("\n # Determination of Cofficient of Friction # \n")
+
+IDf=sqrt((ODf^2)-((((ODi^2)-(IDi^2))*hi)/hf)) //new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi^2-IDi^2)*hi=(3.14/4)*(ODf^2-IDf^2)*hf
+ID=((IDi-IDf)/IDi)*100//change in internal diameter
+
+printf("\n\n With a 50 percent reduction in height and a %d reduction in internal diameter, from the book data Cofficient of Friction = 0.21",ID)
+
+
+
+
+
diff --git a/3308/CH32/EX32.1/Ex32_1.txt b/3308/CH32/EX32.1/Ex32_1.txt new file mode 100755 index 000000000..a87bffe7b --- /dev/null +++ b/3308/CH32/EX32.1/Ex32_1.txt @@ -0,0 +1,4 @@ + # Determination of Cofficient of Friction #
+
+
+ With a 50 percent reduction in height and a 35 reduction in internal diameter, from the book data Cofficient of Friction = 0.21
\ No newline at end of file diff --git a/3308/CH36/EX36.1/Ex36_1.sce b/3308/CH36/EX36.1/Ex36_1.sce new file mode 100755 index 000000000..bb2622356 --- /dev/null +++ b/3308/CH36/EX36.1/Ex36_1.sce @@ -0,0 +1,32 @@ +
+clc
+// Given that
+T=2.6//in mm wall thickness
+USL=3.2//in mm upper specification limit
+LSL=2//in mm lower specification limit
+Y=2.6//in mm mean
+s=0.2//in mm standard deviation
+C1=10//in dollar shipping included cost
+C2=50000//in dollars improvement cost
+n=10000//sections of tube per month
+// Sample Problem on page no. 978
+
+printf("\n # Production of Polymer Tubing # \n")
+
+k=C1/(USL-T)^2
+LossCost=k*(((Y-T)^2)+(s^2))
+//after improvement the variation is half
+s1=0.2/2
+LossCost1=k*(((Y-T)^2)+(s1^2))
+printf("\n\n Taguchi Loss Function = $ %f per unit ",LossCost1)
+//answer in the book is approximated to $0.28 per unit
+
+savings=(LossCost-LossCost1)*n
+paybackperiod=C2/savings
+printf("\n\n Payback Period = %f months",paybackperiod)
+//answer in the book is 6.02 months due to approximation savings
+
+
+
+
+
diff --git a/3308/CH36/EX36.1/Ex36_1.txt b/3308/CH36/EX36.1/Ex36_1.txt new file mode 100755 index 000000000..c9cd91b76 --- /dev/null +++ b/3308/CH36/EX36.1/Ex36_1.txt @@ -0,0 +1,7 @@ +
+ # Production of Polymer Tubing #
+
+
+ Taguchi Loss Function = $ 0.277778 per unit
+
+ Payback Period = 6.000000 months
\ No newline at end of file diff --git a/3308/CH36/EX36.2/Ex36_2.sce b/3308/CH36/EX36.2/Ex36_2.sce new file mode 100755 index 000000000..fbd5d19a5 --- /dev/null +++ b/3308/CH36/EX36.2/Ex36_2.sce @@ -0,0 +1,39 @@ +clc
+// Given that
+n=5// in inch sample size
+m=10// in inch number of samples
+// The table of the queston is given of page no.990 Table 36.3
+
+// Sample Problem on page no. 990
+
+printf("\n # Calculation of Control Limits and Standard Deviation# \n")
+avgx=44.296 //from the table 36.3 by adding values of mean of x
+x = avgx/m
+avgR=1.03 //from the table 36.3 by adding values of R
+R = avgR/m
+//from the data in the book
+A2=0.577
+D4=2.115
+D3=0
+UCLx = x+(A2*R)
+LCLx = x-(A2*R)
+printf("\n\n Control Limits for Averages are =\n UCLx = %f in \n UCLy = %f in",UCLx,LCLx)
+
+UCLR =D3*R
+LCLR =D4*R
+
+printf("\n\n Control Limits for Ranges are =\n UCLR = %f in \n UCLR = %f in",UCLR,LCLR)
+
+//from table
+d2=2.326
+sigma= R/d2
+printf("\n\n Standard Deviation = %f in",sigma)
+
+
+
+
+
+
+
+
+
diff --git a/3308/CH36/EX36.2/Ex36_2.txt b/3308/CH36/EX36.2/Ex36_2.txt new file mode 100755 index 000000000..60e1e3a34 --- /dev/null +++ b/3308/CH36/EX36.2/Ex36_2.txt @@ -0,0 +1,12 @@ +# Calculation of Control Limits and Standard Deviation#
+
+
+ Control Limits for Averages are =
+ UCLx = 4.489031 in
+ UCLy = 4.370169 in
+
+ Control Limits for Ranges are =
+ UCLR = 0.000000 in
+ UCLR = 0.217845 in
+
+ Standard Deviation = 0.044282 in
\ No newline at end of file diff --git a/3308/CH9/EX9.1/Ex9_1.sce b/3308/CH9/EX9.1/Ex9_1.sce new file mode 100755 index 000000000..84b918128 --- /dev/null +++ b/3308/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,21 @@ +clc +// Given that +x=0.2// Area fraction of the fibre in the composite +Ef= 300 // Elastic modulus of the fibre in GPa +Em= 100 // Elastic modulus of the matrix in GPa + +// Sample Problem on page no. 229 + +printf("\n # application of reinforced plastics # \n") + +Ec = x*Ef + (1-x)*Em +printf("\n\n The Elastic Modulus of the composite is = %d GPa",Ec) + +//Let Pf/Pm be r +r=x*Ef/((1-x)*Em) + +//Let Pc/Pf be R +R=1+(1/r) // from the relation Pc = Pf + Pm +P=(1*100)/R +printf("\n\n The Fraction of load supported by Fibre is = %f Percent",P) +// Answer in the book is approximated to 43 % diff --git a/3308/CH9/EX9.1/Ex9_1.txt b/3308/CH9/EX9.1/Ex9_1.txt new file mode 100755 index 000000000..2e6d0af07 --- /dev/null +++ b/3308/CH9/EX9.1/Ex9_1.txt @@ -0,0 +1,6 @@ + # application of reinforced plastics #
+
+
+ The Elastic Modulus of the composite is = 140 GPa
+
+ The Fraction of load supported by Fibre is = 42.857143 Percent
\ No newline at end of file |