initial commit / add all books

11 files changed, 277 insertions, 0 deletions

diff --git a/1205/CH3/EX3.1/S_3_1.sce b/1205/CH3/EX3.1/S_3_1.sce
new file mode 100644
index 000000000..3991497ce
--- /dev/null
+++ b/1205/CH3/EX3.1/S_3_1.sce
@@ -0,0 +1,35 @@
+clc;

+// Given data

+F=500; // N , Vertical force applied to end of lever

+theta=60;// degree, angle made by lever with +ve X axis

+theta=theta*%pi/180;// Conversion of angle into radian

+l=600; // mm , length of lever

+

+// a ) Momemt about O

+d=l*cos(theta);// mm ,perpendicular distance from  o to the line of action

+d=d/1000; // m, conversion into meter

+Mo=F*d;// N.m, Magnitude of moment about O 

+printf("Magnitude of moment about O of the 500 N is %.2f N.m and it is in clockwise direction as force tends to rotate lever clockwise\n",Mo);

+

+// b) Horizontal force

+

+d=l*sin(theta);//mm, perpendicular distance from  o to the line of action

+d=d/1000; // m, conversion into meter

+F=Mo/d;// N, Horizontal Force at A required to produce same Moment about O

+printf("Magnitude of Horizontal Force at A required to produce same Moment about O is %.2f N\n",F);

+

+// c)Smallest force

+

+// F is smaller when d is maximum in expression Mo=F*d, so we choose force perpendicular to OA

+d=0.6;// m ,perpendicular distance from  o to the line of action

+F=Mo/d;// N, Smallest Force at A required to produce same Moment about O

+printf("Magnitude of smallest Force at A required to produce same Moment about O is %.2f N\n",F);

+

+//d) 1200 N vertical force

+F=1200;// N, verical force producing same movement on lever acting at pt B

+d=Mo/F;// m, perpendicular distance from  o to the line of action of force

+OB=d/cos(theta);//m, distance of point B from O

+OB=OB*1000;//mm, conversion into millimeter

+printf("Verical force of 1200 N must act at %.2f mm far from the shaft to create same moment about O\n",OB);

+

+

diff --git a/1205/CH3/EX3.10/S_3_10.sce b/1205/CH3/EX3.10/S_3_10.sce
new file mode 100644
index 000000000..382c4acba
--- /dev/null
+++ b/1205/CH3/EX3.10/S_3_10.sce
@@ -0,0 +1,40 @@
+clc;

+BE=[75,-150,50];//mm position vector BE

+F_B=700;//N*lambada_BE 

+lambda_BE=BE/norm(BE);//mm,Unit vector

+lambda_BE=lambda_BE/1000;//m, Converting into meter

+

+// Using meters and newton

+r_BA=[0.075,0,0.05];//m

+r_CA=[0.075,0,-0.05];//m

+r_DA=[0.1,-0.1,0];//m

+F_B=[300,-600,200];//N

+F_C=[707,0,-707];//N

+F_D=[600,1039,0];//N

+

+//R=sum( F)

+

+R=F_B+F_C+F_D;//N , Resultant force

+printf("Resultant force can be shown as R= (%.2f N)i +(%.2f N)j +(%.2f N)k \n",R(1),R(2),R(3));

+

+// Componenets of moment M_BA along X,Y and Z direction respectively

+M_BAx=det([r_BA(2),r_BA(3); F_B(2), F_B(3)]);//N.m

+M_BAy=-det([r_BA(1),r_BA(3) ; F_B(1),F_B(3)]);//N.m

+M_BAz=det([r_BA(1),r_BA(2) ;F_B(1), F_B(2)]);// N.m 

+M_BA=[M_BAx,M_BAy,M_BAz];

+

+// Componenets of moment M_CA along X,Y and Z direction respectively

+M_CAx=det([r_CA(2),r_CA(3); F_C(2), F_C(3)]);//N.m

+M_CAy=-det([r_CA(1),r_CA(3) ; F_C(1),F_C(3)]);//N.m

+M_CAz=det([r_CA(1),r_CA(2) ;F_C(1), F_C(2)]);// N.m 

+M_CA=[M_CAx,M_CAy,M_CAz];

+

+// Componenets of moment M_DA along X,Y and Z direction respectively

+M_DAx=det([r_DA(2),r_DA(3); F_D(2), F_D(3)]);//N.m

+M_DAy=-det([r_DA(1),r_DA(3) ; F_D(1),F_D(3)]);//N.m

+M_DAz=det([r_DA(1),r_DA(2) ;F_D(1), F_D(2)]);// N.m 

+M_DA=[M_DAx,M_DAy,M_DAz];

+

+//M_RA=sum(r*F)

+M_RA=M_BA+M_CA+M_DA;//N.m

+printf("Resultant moment can be shown as M_RA= (%.2f N)i +(%.2f N)j +(%.2f N)k ",M_RA(1),M_RA(2),M_RA(3));

diff --git a/1205/CH3/EX3.11/S_3_11.sce b/1205/CH3/EX3.11/S_3_11.sce
new file mode 100644
index 000000000..ad0802b27
--- /dev/null
+++ b/1205/CH3/EX3.11/S_3_11.sce
@@ -0,0 +1,30 @@
+clc;

+

+//Cross product i*j=k, k*j=-i, j*j=0.......1

+//Let say cross product r*f as M

+r1=[0,0,0];//m

+F1=-180;//N, j

+M1=[0,0,0];//kN.m

+

+r2=[3,0,0];//m, i

+F2=-54;//N j

+M2=[0,0,r2(1)*F2];//kN k

+

+r3=[3,0,1.5];//m

+F3=-36;//N,j

+M3=[-r3(3)*F3,0,r3(1)*F3];//kN.m

+

+r4=[1.2,0,3];//m

+F4=-90;//N,j

+M4=[-r4(3)*F4,0,r4(1)*F4];//kN.m

+

+R=F1+F2+F3+F4;//kN, resultant force

+M_RO=M1+M2+M3+M4;//kN.m 

+

+//r*R=M_RO

+//r=xi+zk

+//R*x k-R*z i=M_RO

+x=M_RO(3)/R;//m,

+z=-M_RO(1)/R;//m

+

+printf("Resultant of given system of forces is R= %.2f kN at x= %.2f m, z=%.2f m \n",R,x,z);

diff --git a/1205/CH3/EX3.2/S_3_2.sce b/1205/CH3/EX3.2/S_3_2.sce
new file mode 100644
index 000000000..f7463f0a3
--- /dev/null
+++ b/1205/CH3/EX3.2/S_3_2.sce
@@ -0,0 +1,14 @@
+clc;

+// Given data

+F=800; // N , Force applied on bracket

+theta=60;// degree, angle made by lever with +ve X axis

+theta=theta*%pi/180;// Conversion of angle into radian

+r_AB=[-0.2, 0.16];//m vector drawn from B to A resolved in rectangular component

+F=[F*cos(theta), F*sin(theta)]//N , vector F resolved in rectangular component 

+k=1;// Unit vector along Z axis 

+

+// M_B=r_AB * F relation 3.7 from section 3.5

+M_B=det([r_AB; F])*k;// N.m 

+printf("The moment of force 800 N about B is %.2f N.m . -ve sign shows its acting clockwise\n",M_B);

+

+

diff --git a/1205/CH3/EX3.3/S_3_3.sce b/1205/CH3/EX3.3/S_3_3.sce
new file mode 100644
index 000000000..8dc4bbd80
--- /dev/null
+++ b/1205/CH3/EX3.3/S_3_3.sce
@@ -0,0 +1,16 @@
+clc;

+// Given data

+P=40; // N , Force applied to shift lever 

+alpha=25;// degree, angle made by force P with -ve X axis

+alpha=alpha*%pi/180;// Conversion of angle into radian

+x=0.2;//m , Hirizontal distance of A from B

+y=0.6;//m, Vertical distance of A from B

+

+

+Px=P*cos(theta);// N , horizontal component

+Py=P*sin(theta);//N , Vertical component

+

+M_B=-x*Px-y*Py//N.m , here negative signs are taken as each component creates moment clockwise

+printf("The moment of force P about B is %.2f N.m . -ve sign shows its acting clockwise\n",M_B);

+

+

diff --git a/1205/CH3/EX3.4/S_3_4.sce b/1205/CH3/EX3.4/S_3_4.sce
new file mode 100644
index 000000000..11d804457
--- /dev/null
+++ b/1205/CH3/EX3.4/S_3_4.sce
@@ -0,0 +1,24 @@
+clc;

+// Given data

+// M_A=r_CA * F relation 3.7 from section 3.5

+f=200; // N , Magnitude of Force directed along CD

+r_CA=[0.3,0, 0.08];//m, vector AC reprecsented in rectangular component

+//lambda=CD/norm(CD)-m, Unit vector along CD

+//F=f*lambda;//m, Force 

+CD=[-0.3, 0.24, -0.32];//Vector CD resolved into rectangular component

+// norm(CD); m, magnitude of vector CD

+

+lambda=CD/norm(CD);//m, Unit vector along CD

+F=f*lambda;//m, Force 

+// M_A=r_CA * F relation 3.7 from section 3.5

+//i=1; j=1; k=1; Unit vectors along X, Y and Z direction respectively

+

+// Componenets of moment M_A along X,Y and Z direction respectively

+M_Ax=det([r_CA(2),r_CA(3); F(2), F(3)]);//N.m

+M_Ay=-det([r_CA(1),r_CA(3) ; F(1),F(3)]);//N.m

+M_Az=det([r_CA(1),r_CA(2) ;F(1), F(2)]);// N.m 

+

+printf("Answer can be written as M_B = %.2f N.m i + %.2f N.m j + %.2f N.m k \n",M_Ax,M_Ay,M_Az);

+

+

+

diff --git a/1205/CH3/EX3.5/S_3_5.sce b/1205/CH3/EX3.5/S_3_5.sce
new file mode 100644
index 000000000..485f608b5
--- /dev/null
+++ b/1205/CH3/EX3.5/S_3_5.sce
@@ -0,0 +1,32 @@
+clc;

+

+// Mo=r_BO * F_B relation 3.7 from section 3.5

+r_BO=[0,7,0];//m

+//F_B=T_AB+T_BC; N , 

+mT_AB=555;//N, Tension in Cable AB

+mT_BC=660;//N, Tension in cable AC

+CA=[0.3,0, 0.08];//m, vector AC reprecsented in rectangular component

+//lambda=CD/norm(CD)-m, Unit vector along CD

+//F=f*lambda;//m, Force 

+BA=[-0.75,-7,6];//m, Position vector BA resolved into rectangular component

+BC=[4.25,-7,1];//m,Position vector BC resolved into rectangular component

+

+lambda_BA=BA/norm(BA);//m, Unit vector along BA

+T_AB=mT_AB*lambda_BA;//m, Force along cable AB

+

+lambda_BC=BC/norm(BC);//m, Unit vector along Bc

+T_BC=mT_BC*lambda_BC;//m, Force along cable BC

+

+F_B=T_AB+T_BC;// N

+// M_A=r_CA * F relation 3.7 from section 3.5

+//i=1; j=1; k=1; Unit vectors along X, Y and Z direction respectively

+

+// Componenets of moment M_A along X,Y and Z direction respectively

+M_Ax=det([r_BO(2),r_BO(3); F_B(2), F_B(3)]);//N.m

+M_Ay=-det([r_BO(1),r_BO(3) ; F_B(1),F_B(3)]);//N.m

+M_Az=det([r_BO(1),r_BO(2) ;F_B(1), F_B(2)]);// N.m 

+

+printf("Answer can  be written as M_B = %.2f N.m i + %.2f N.m j + %.2f N.m k \n",M_Ax,M_Ay,M_Az);

+

+

+

diff --git a/1205/CH3/EX3.6/S_3_6.sce b/1205/CH3/EX3.6/S_3_6.sce
new file mode 100644
index 000000000..644bfd191
--- /dev/null
+++ b/1205/CH3/EX3.6/S_3_6.sce
@@ -0,0 +1,17 @@
+clc;

+//Given data

+// Moment arms

+x=0.45;//m

+y=0.30;//m

+z=0.23;//m

+

+//couple Forces 

+Fx=-150;//N

+Fy=100;//N

+Fz=100;//N

+

+Mx=Fx*x;//N.m, Component of Moment along X axis

+My=Fy*y;//N.m, Component of Moment along Y axis

+Mz=Fz*z;//N.m, Component of Moment along Z axis

+//This three moments represent component of single couple M 

+printf("Couple M equivalent to two couple can be written as M = %.2f N.m i + %.2f N.m j + %.2f N.m k \n",Mx,My,Mz);

diff --git a/1205/CH3/EX3.7/S_3_7.sce b/1205/CH3/EX3.7/S_3_7.sce
new file mode 100644
index 000000000..325348336
--- /dev/null
+++ b/1205/CH3/EX3.7/S_3_7.sce
@@ -0,0 +1,12 @@
+clc;

+Mo=-24;//N.m *k, Couple of moment 

+f=400;//N, Magnitude of force

+OB=300;//mm,Distance of force from point O

+theta=60;// degree, angle made by lever with +ve X axis

+theta=theta*%pi/180;// Conversion of angle into radian

+// Mo=BC_ * F relation 3.7 from section 3.5

+//BC_ * F=BC*f*cos(theta)------ Defination of cross product

+BC=Mo/(-cos(theta)*f);//m 

+BC=BC*1000;//mm, Conversion into millimeter

+OC=OB+BC;//mm, Distance from the shaft to the point of application of this equivalenet force

+printf("Distance from the shaft to the point of application of this equivalenet single force is %.2f mm",OC)

diff --git a/1205/CH3/EX3.8/S_3_8.sce b/1205/CH3/EX3.8/S_3_8.sce
new file mode 100644
index 000000000..2dfa70479
--- /dev/null
+++ b/1205/CH3/EX3.8/S_3_8.sce
@@ -0,0 +1,24 @@
+clc;

+

+

+// Given forces with direction shown by sign -ve to downwards, +ve for upwards. j shows that firceis along y direction

+f1=150;//N, j 

+f2=-600;//N, j

+f3=100;//N , j

+f4=-250;//N,j

+//a) force couple system at A

+R=f1+f2+f3+f4;//N, j Resultant force, sum of all forces

+//M_RA=sum(r*f)

+M_RA=1.6*f2+2.8*f3+4.8*f4;//m, k sum of moments by each force

+printf("Equivalent force couple system at A is thus R= %.2f N and M_RA= %.2f N.m \n",R,M_RA);

+//B) Force couple system at B

+BA=-4.8;//m, i

+M_RB=M_RA+BA*R;//N.m

+printf("Equivalent force couple system at B is thus R= %.2f N and M_RB= %.2f N.m \n",R,M_RB);

+

+//c)single force or resultant

+// r*R=M_RA

+

+//x.i * (-600N)j=-(1800N.m)k

+x=M_RA/R;//m, distance of point of application from A

+printf("Equivalent single force is defined as at  R= %.2f N and acts at x= %.2f m \n",R,x);

diff --git a/1205/CH3/EX3.9/S_3_9.sce b/1205/CH3/EX3.9/S_3_9.sce
new file mode 100644
index 000000000..51a34d218
--- /dev/null
+++ b/1205/CH3/EX3.9/S_3_9.sce
@@ -0,0 +1,33 @@
+clc;

+// Resolving forces 125 N and 90  N

+f1=125;//N

+f2=90;//N

+f3=200;//N

+theta=45;// degree, angle made by f1 with x axis

+theta=theta*%pi/180;// Conversion of angle into radian

+alpha=30;// degree, angle made by force f2 with y axis

+alpha=alpha*%pi/180;// Conversion of angle into radian

+f1x=f1*cos(theta);//N, X component of 125 N.

+f1y=-f1*sin(theta);//N, Y component of 125 N

+

+f2x=-f2*sin(alpha);//N, X component of 90N

+f2y=-f2*cos(alpha);//N, Y component of 90N

+

+// Equivalent force at A

+Rx=f1x+f2x;//N

+Ry=f1y+f2y-f3;//N

+R=[Rx,Ry];//N

+R=norm(R);//N, magnitude of resultant

+theta=atan(Ry/Rx);//radian, angle of resultant with X axis

+theta=theta*180/%pi;

+

+printf("Equivalent force at A is R=%.2fN with angle %.2f Degrees \n ",R,theta);

+

+//equivalent couple at A

+//Clockwise moments are negative

+Meq=-550*f2*sin(35*%pi/180)-800*f3*sin(65*%pi/180)-1200*f1*sin(65*%pi/180);//N.mm , sum of all moments

+Meq=Meq/1000;// N.m conversion into N.m

+

+

+printf("Equivalent couple at A is Meq= %.2f N.m \n",Meq);

+