initial commit / add all books

7 files changed, 209 insertions, 0 deletions

diff --git a/1205/CH6/EX6.1/S_6_1.sce b/1205/CH6/EX6.1/S_6_1.sce
new file mode 100644
index 000000000..c8a9c9bb3
--- /dev/null
+++ b/1205/CH6/EX6.1/S_6_1.sce
@@ -0,0 +1,33 @@
+clc;

+//Entire truss

+//Applying sum(M_C)=0

+E=(10*12+5*6)/3;//kN

+

+//Applying sum Fx=0

+Cx=0

+

+//Applying sumFy=0

+

+Cy=10+5-E;//kN

+

+//At joint A

+//By proportion 10kN/4=F_AB/3=F_AD/5

+F_AB=10/4*3;//kN, force in member AB

+F_DA=10/4*5;//kN, force in member AD

+

+//At joint D

+F_DB=F_DA;//kN, force in member DB

+F_DE=2*3/5*F_DA;//kN, force in member DE

+

+//At joint B

+//applying sumFy=0

+F_BE=5/4*(-5-4/5*F_DB);//kN, force in member BE

+//Applying sumFx=0

+

+F_BC=F_AB+3/5*F_DB-3/5*F_BE;//kN, force in member BC

+

+//At joint E

+//Applying sumFx=0

+F_EC=-5/3*(F_DE-3/5*F_BE);//kN, Force in member EC

+

+printf("The forces in member of truss are \n F_AB= %.1f kN T \n F_AD= %.1f kN C, \n F_DB= %.1f kN T, \n F_DE= %.0f kN C \n F_BE= %.2f kN  \n F_BC= %.2f kN \n F_EC= %.2f kN ",F_AB,F_DA,F_DB,F_DE,F_BE,F_BC,F_EC);

diff --git a/1205/CH6/EX6.2/S_6_2.sce b/1205/CH6/EX6.2/S_6_2.sce
new file mode 100644
index 000000000..d2197ba4c
--- /dev/null
+++ b/1205/CH6/EX6.2/S_6_2.sce
@@ -0,0 +1,23 @@
+clc;

+//Entire truss

+v1=140;//kn, verical force 1

+v2=140;//kN, Vertical force 2

+h=80;//kN , Horizontal force

+//Applying sum(M_B)=0

+J=(v1*4+v2*12+h*5)/16;//kN

+

+//Applying sum Fx=0

+Bx=-h;//kN, negative sign shows it is along negative x axis

+

+//Applying sumFy=0

+

+By=v1+v2-J;//kN

+

+//Force in member EF

+//Applying sumFy=0

+F_EF=By-v2;//kN, Force in member EF

+printf("Force in member EF is %.0f kN \n Negative sign shows member is in compression \n",F_EF);

+

+//Force in member GI

+F_GI=(-J*4-Bx*5)/5;//kN Force in member GI

+printf("Force in member GI is %.0f kN \n Negative sign shows member is in compression \n",F_GI);

diff --git a/1205/CH6/EX6.3/S_6_3.sce b/1205/CH6/EX6.3/S_6_3.sce
new file mode 100644
index 000000000..59849fddb
--- /dev/null
+++ b/1205/CH6/EX6.3/S_6_3.sce
@@ -0,0 +1,37 @@
+clc;

+//Entire truss

+vB=1;//kN, verical force at B

+vD=1;//kN, verical force at D

+vF=1;//kN, verical force at F

+vH=1;//kN, verical force at H

+vJ=1;//kN, verical force at J

+vC=5;//kN, verical force at C

+vE=5;//kN, verical force at E

+vG=5;//kN, verical force at G

+h=8;//m, height

+v=5;//m, horizontal distance between successive node

+

+A=12.50;//kN, reaction at A

+L=7.50;//kN, reaction at L

+

+alpha=atan(h/3/v);// rad, angle made by inclined members with  X axis

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

+

+

+

+//Force in member GI

+//Applying sum(M_H)=0

+F_GI=(L*2*v-vJ*v)/(2*v*tan(alpha));//kN Force in member GI

+printf("Force in member GI is %.2f kN \n ",F_GI);

+

+//Force in member FH

+//Applying sum(M_G)=0

+F_FH=(L*3*v-vH*v-vJ*2*v)/(-h*cos(alpha));//kN, Force in member FH

+printf("Force in member FH is %.2f kN \n Negative sign shows member is in compression \n",F_FH);

+

+

+//Force in member GH

+be=atan(v/(2*v*tan(alpha)));//rad, as tan(be)=GI/HI

+//Applying sum(M_L)=0

+F_GH=(-vH*v-vJ*2*v)/(3*v*cos(be));//kN, Force in member FH

+printf("Force in member GH is %.3f kN \n Negative sign shows member is in compression \n",F_GH);

diff --git a/1205/CH6/EX6.4/S_6_4.sce b/1205/CH6/EX6.4/S_6_4.sce
new file mode 100644
index 000000000..0a3cfcaee
--- /dev/null
+++ b/1205/CH6/EX6.4/S_6_4.sce
@@ -0,0 +1,23 @@
+clc;

+//Entire truss

+//Applying sum(Fy)=0

+Ay=480;//N, Y component of reaction at A

+//Applying sum(M_A)=0

+B=480*100/160;//N,  reaction at B

+//Applying sum(Fx)=0

+Ax=-300;//N, X component of reaction at A

+

+alpha=atan(80/150);//radian

+

+//Free body member BCD

+

+//Applying sum(M_C)=0

+F_DE=(-480*100-B*60)/(sin(alpha)*250);//N,  Force in link DE

+printf("Force in link DE is F_DE=%.0f N\n Negative sign shows force is compressive\n",F_DE);

+//Applying sum(Fx)=0

+Cx=F_DE*cos(alpha)-B;//N, X component of force exerted at C

+//Applying sum(Fy)=0

+Cy=F_DE*sin(alpha)+Ay;//N, Y component of force exerted at C

+printf("Components of force exerted at C is Cx=%.0f N and Cy=%.0f N \n",Cx,Cy);

+

+

diff --git a/1205/CH6/EX6.5/S_6_5.sce b/1205/CH6/EX6.5/S_6_5.sce
new file mode 100644
index 000000000..67fb998ee
--- /dev/null
+++ b/1205/CH6/EX6.5/S_6_5.sce
@@ -0,0 +1,38 @@
+clc;

+P=18;//kN, Force applied at D

+AF=3.6;//m, Length AF

+EF=2;//m, Length EF

+ED=2;//m, Length ED

+DC=2;//m, Length DC

+//Entire frame

+//Applying sum(M_F)=0

+Ay=-P*(EF+ED)/AF;//kN, Y component of reaction at A

+

+//Applying sum(Fx)=0

+Ax=-P;//kN, X component of reaction at A

+//Applying sum(Fy)=0

+F=-Ay;//kN,  reaction at B

+

+

+printf("Components of force exerted at A is Ax=%.0f kN and Ay=%.0f kN \n",Ax,Ay);

+printf("Force exerted at F is F=%.0f kN \n",F);

+//Free body member BE

+//Applying sum(Fx)=0

+//B=E, and as it is 2 force member

+By=0;

+Ey=0;

+

+//Member ABC

+//Applying sum(Fy)=0

+Cy=-Ay;//kN, Y component of force exerted at C

+//Applying sum(M_C)=0

+B=(Ay*AF-Ax*(DC+ED+EF))/(ED+DC);//kN,  Force in link DE

+printf("Force exerted at B is B=%.0f kN  \n",B);

+//Applying sum(Fx)=0

+Cx=-Ax-B;//kN, X component of force exerted at C

+

+printf("Components of force exerted at C is Cx=%.0f kN and Cy=%.0f kN \n",Cx,Cy);

+

+printf("Negative signs shows forces are in negative direction\n")

+

+

diff --git a/1205/CH6/EX6.6/S_6_6.sce b/1205/CH6/EX6.6/S_6_6.sce
new file mode 100644
index 000000000..8adfcdbd3
--- /dev/null
+++ b/1205/CH6/EX6.6/S_6_6.sce
@@ -0,0 +1,34 @@
+clc;

+P=3;//kN, Horizontal Force applied at A

+AB=1;//m, perpendicular distance between A and B

+BD=1;//m, perpendicular distance between D and B

+CD=1;//m, perpendicular distance between C and D

+FC=1;//m, perpendicular distance between C and F

+EF=2.4;//m, perpendicular distance between E and F

+//Entire frame

+//Applying sum(M_E)=0

+Fy=P*(AB+BD+CD+FC)/EF;//kN, Y component of reaction at F

+

+

+//Applying sum(Fy)=0

+Ey=-Fy;//kN, Y component of  reaction at E

+

+//Free body member ACE

+//Applying sum(Fy)=0, and sum(M_E)=0 we get 2 equation

+A=[-AB/sqrt(AB^2+EF^2),CD/sqrt(CD^2+EF^2);-EF/sqrt(AB^2+EF^2)*(AB+BD+CD+FC),-EF/sqrt(CD^2+EF^2)];// Matrix of coefficients

+B=[Ey;-P*(AB+BD+CD+FC)];// Matrix B

+X=linsolve(A,B);//kN Solution matrix

+F_AB=X(1);//kN, Forec inmember AB

+F_CD=X(2);//kN, Forec inmember CD

+Ex=-P-EF/sqrt(AB^2+EF^2)*F_AB-EF/sqrt(CD^2+EF^2)*F_CD;//kN, X component of force exerted at E

+//Free body : Entire frame

+//Applying sum(F_X)=0

+Fx=-P-Ex;//kN, X component of force exetered at F

+printf("Components of force exerted at F is Fx=%.1f kN and Fy=%.0f kN \n",Fx,Fy);

+printf("Force in member AB is F_AB=%.1f kN \n",F_AB);

+printf("Force in member CD is F_CD=%.1f kN \n",F_CD);

+printf("Components of force exerted at E is Ex=%.1f kN and Ey=%.1f kN \n",Ex,Ey);

+

+printf("Negative signs shows forces are in negative direction\n")

+

+

diff --git a/1205/CH6/EX6.7/S_6_7.sce b/1205/CH6/EX6.7/S_6_7.sce
new file mode 100644
index 000000000..8fa50ec95
--- /dev/null
+++ b/1205/CH6/EX6.7/S_6_7.sce
@@ -0,0 +1,21 @@
+clc;

+//Part 1

+printf("By theoritical calculation we getb F_DH=W*cos(theta)/sin(theta)\n");

+printf("Therefore result obtained is independant of d\n");

+//Given data

+

+a=0.70;//m

+theta=60;//degree

+theta=theta*%pi/180;//radian, conversion into radian

+L=3.20;//m

+m=1000;//kg, mass of crate

+g=9.81;//m/s^2, acceleration due to gravity

+

+DH=sqrt(a^2+L^2-2*a*L*cos(theta));//m, by cosine rule

+W=m*g;//N, wight of crate

+W=W/1000;//kN, conversion into kN

+//Recalling equation got from theoritical part

+EH=L;//m

+F_DH=W*DH/EH/tan(theta);//kN, Force exerted by cylinder

+

+printf("Force exerted by each cylinder is F_DH=%.2f kN \n",F_DH);