initial commit / add all books

7 files changed, 166 insertions, 0 deletions

diff --git a/1538/CH5/EX5.1/Ex5_1.sce b/1538/CH5/EX5.1/Ex5_1.sce
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+//example-5.1

+//page no-136

+//given

+//wavelength of X-rays beams of light

+lambda=0.824*10^-10  //m

+//glancing angle of the incident light

+theta1=(8+35/60)*(%pi)/180  //radians

+n1=1

+//to find theta3 i.e at

+n3=3

+//as we know that

+//2*d*sin(theta)=n*lambda

+//so for n1 and n3 we get in the same way and solving together we get 

+theta3=asin(3*sin(theta1))

+//so 

+d=lambda/2/sin(theta1)

+printf ("the galncing angle for thethird order diffraction is and interplanar spacing of the crystal is 2.761 A")

diff --git a/1538/CH5/EX5.2/Ex5_2.sce b/1538/CH5/EX5.2/Ex5_2.sce
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+//example-5.2

+//page no-141

+//given

+//bragg's angle of reflection

+theta1=17.03*(%pi)/180  //radians

+//wavelength of light

+lambda=0.71  //A

+//according to bragg's equation 

+//n*lambda=2*d*sin(theta)

+//for n=1

+d=lambda/2/sin(theta1)  //A

+//given that h^2+k^2+l^2=8

+//let (h^2+k^2+l^2)^1/2=H

+//we get

+H=sqrt(8)

+a=d*H  //A

+//since h^2+k^2+l^2=8 ,hence the reflecting planes will be (220). family of planes (220) include (220), (202), (022) ,etc.

diff --git a/1538/CH5/EX5.3/Ex5_3.sce b/1538/CH5/EX5.3/Ex5_3.sce
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+//example-5.3

+//page n0-141

+//given

+//lattice constant

+a=1.54  //A

+//wavelength of beam of light

+lambda=1.54 //A

+//according to bragg's equation

+//n*lambda=2*d*sin(theta)

+//following angles are given

+theta1=20.3*(%pi)/180

+theta2=29.2*(%pi)/180

+theta3=36.7*(%pi)/180

+theta4=43.6*(%pi)/180

+//interplaner spadcing is 

+d1=lambda/(2*sin(theta1))  //A

+d2=lambda/(2*sin(theta2))  //A

+d3=lambda/(2*sin(theta3))  //A

+d4=lambda/(2*sin(theta4))  //A

+//magnitude of bragg's 

+//we have h^2+k^2+l^2=a^2/d^2

+//let a^2/d^2= D for notation only

+//so

+D1=2

+D2=4

+D3=6

+D4=8

+//so from bragg's magnitude we can get (hkl)

+//(hkl1)=(110)

+//(hkl3)=(200)

+//(hkl3)=(211)

+//(hkl4)=(220)

+printf ("the reflection will take from {110},{200},{211} and (220)")

diff --git a/1538/CH5/EX5.5/Ex5_5.sce b/1538/CH5/EX5.5/Ex5_5.sce
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+//example-5.5

+//page no-146

+//given

+//wavelength of X-ray

+lambda=1.54  //A

+//diameter of powder camera

+D=114.6  //mm

+//radius of powder camera

+R=D/2  //mm

+//value of l

+l1=86

+l2=100

+l3=148

+l4=180

+l5=188

+l6=232

+l7=272

+//we know that

+//theta=l/4

+//so

+theta1=l1/4*(%pi)/180   //radians

+theta2=l2/4*(%pi)/180   //radians

+theta3=l3/4*(%pi)/180   //radians

+theta4=l4/4*(%pi)/180   //radians

+theta5=l5/4*(%pi)/180   //radians

+theta6=l6/4*(%pi)/180   //radians

+theta7=l7/4*(%pi)/180   //radians

+//now values of sin (theta) and sin(theta2)

+S1=sin(theta1)

+SS1=(sin(theta1))^2

+S2=sin(theta1)

+SS2=(sin(theta1))^2

+S3=sin(theta1)

+SS3=(sin(theta1))^2

+S4=sin(theta1)

+SS4=(sin(theta1))^2

+S5=sin(theta1)

+SS5=(sin(theta1))^2

+S6=sin(theta1)

+SS6=(sin(theta1))^2

+S7=sin(theta1)

+SS7=(sin(theta1))^2

+//so the ratio can be expressed as

+//3:4:8:11:12:16:19

+printf ("from the extinction rule, we notice that this is an FCC Structure")

+//the lattice parameter for highest bragg's angle is

+//a=lambda*sqrt(h^2+k^2+l^2)/(2*sin(theta))

+//here h^2+k^2+l^2=19

+//and let h^2+k^2+l^2 =M for notation

+M=19

+a=lambda*sqrt(M)/(2*sin(theta6))  //A

+printf ("lattice parameter of material is %f A",a)

diff --git a/1538/CH5/EX5.6/Ex5_6.sce b/1538/CH5/EX5.6/Ex5_6.sce
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index 000000000..c77a09d18
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+//example-5.6

+//page no-158

+//given

+//ASTM number

+n=12

+//as we know that the number of grains N observed on photomicrograph is given by

+N=2^(n-1)

+//as we know that grain size diameter is given by

+d=1/sqrt((N/645)*10^4)  //mm  because 1 square inch=645 mm^2

+printf ("the grain diameter for an ASTM number 12 is %f mm",d)

diff --git a/1538/CH5/EX5.7/Ex5_7.sce b/1538/CH5/EX5.7/Ex5_7.sce
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+//example-5.7

+//page no-158

+//given

+//no of grains within the view of a micrograph

+n1=41

+//no of grains cut by circumference

+n2=42

+//diameter of circular area

+d=1  //inch

+//area

+A=(%pi)/4*d^2  //inch^2

+//the area density of grains

+

+N=(n1+n2/2)/A  //grains/inch^2

+//grain size

+n=log(N)/log(2)+1  

+printf ("the area density of grains is %f grains/inch^2 and grain size is 8",N)

diff --git a/1538/CH5/EX5.8/Ex5_8.sce b/1538/CH5/EX5.8/Ex5_8.sce
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+//example-5.8

+//page no-159

+//given

+//ASTM no of grains

+ASTM=5

+//area density of grains

+N=2^(ASTM-1)  //grains/inch^2 at magnification of 100*

+//as we know that lineal and areal magnification are related as

+//*100 lineal=*10000 areal

+//therefore

+Nnew=N/0.01/0.01  //grains/inch^2 at 1*

+//average area of one grain

+A=1/Nnew*(2.54)^2  //cm^2

+//now 160000 grains/inch^2 of surface is sqrt(160000)=400 grains/inch of length and this is equal to=(400)^3==6.4*10^7 grains/m^3 of volume

+//surface area of each cubic surface

+S=(1/400)^2  //inch^2

+//there are 6 surfaces in accubic grain

+//thus total surface area of each grain

+T=1/2*6*S*(400)^3/2.54  //cm^2 boundary per cubic cm of steel

+printf ("the boumdary area per cubic centimeter of steel is %f cm^2 boundary per cubic centimeter of steel",T)