10 files changed, 205 insertions, 0 deletions
diff --git a/3482/CH4/EX4.1/Ex4_1.sce b/3482/CH4/EX4.1/Ex4_1.sce
new file mode 100644
index 000000000..c7038c9cf
--- /dev/null
+++ b/3482/CH4/EX4.1/Ex4_1.sce
@@ -0,0 +1,23 @@
+clc;
+//page 166
+//Determination of B
+//At equillibrium +sum(M_A)=0
+//B*1.5m-(9.81kN)(2 m)-(23.5 kN)(6 m)=0, B assumed to be in +ve X direction
+B=(9.81*2+23.5*6)/1.5//kN
+printf("B=%.2f kN \n +ve sign shows reaction is directed as assumed ",B);
+//Determination of Ax
+//Sum Fx=0
+//Ax+B=0
+Ax=-B;//kN
+printf("Ax=%.2f kN\n",Ax);
+//Determination of Ay
+//Sum Fy=0
+//Ay-9.81 kN-23.5kN=0
+Ay=9.81+23.5;//kN
+printf("Ay=%.2f kN\n",Ay);
+A=[Ax,Ay];//kN Adding component
+A=norm(A);//Magnitude of force A
+theta=atan(Ay/Ax);//radians
+theta=theta*180/%pi;//degrees, conversion into degrees
+printf("Reaction at A is A=%.2f kN making angle %.2f degrees \n with + ve x axis ",A,theta);
+//Slight variation in the answer because of roundoff error
diff --git a/3482/CH4/EX4.10/Ex4_10.sce b/3482/CH4/EX4.10/Ex4_10.sce
new file mode 100644
index 000000000..acf5dcc9c
--- /dev/null
+++ b/3482/CH4/EX4.10/Ex4_10.sce
@@ -0,0 +1,17 @@
+clc;
+//page 197
+Tmin=300//lb
+AC=[12 12 0]
+w=[0;-450;0]
+x1=AC*w
+disp(x1)
+x=[0 0 x1]
+lambda=[2/3 2/3 -1/3]*[0;0;-x1]
+y=x*lambda
+disp(y)
+
+//Location of G
+//EG and Tmin are having same direction, so their component should be in proportion
+x=-1.8/Tmin(3)*Tmin(1)+1.8;//m, X co-ordinate of G
+y=-1.8/Tmin(3)*Tmin(2)+3.6;//m, Y co-ordinate of G
+printf("Co-ordinates of G are x=%.0f m and y= %.1f m",x,y);
diff --git a/3482/CH4/EX4.2/Ex4_2.sce b/3482/CH4/EX4.2/Ex4_2.sce
new file mode 100644
index 000000000..96e7e9638
--- /dev/null
+++ b/3482/CH4/EX4.2/Ex4_2.sce
@@ -0,0 +1,15 @@
+clc;
+//Page 148
+
+//At equillibrium equations are +-> sum Fx=0, +sum(M_A)=0, +sum(M_B)=0
+//Sum Fx=0 gives
+Bx=0;//kN
+printf("Bx=%.0f kN \n",Bx);
+//+sum(M_A)=0 gives -(70kN)(0.9m)+By(2.7m)-(27kN)(3.3m)-(27kN)(3.9m)=0, B assumed to be in +ve Y direction
+By=(70*0.9+27*3.3+27*3.9)/2.7//kN
+printf("By=%.2f kN  +ve sign shows reaction is directed as assumed \n",By);
+
+//+sum(M_B)=0 gives -A(2.7m)+(70kN)(1.8m)-(27kN)(0.6m)-(27kN)(1.2m)=0, A assumed to be in +ve Y direction
+A=(70*1.8-27*0.6-27*1.2)/2.7//kN
+printf("A=%.2f kN  +ve sign shows reaction is directed as assumed \n",A);
+//Answer displayed in KN
diff --git a/3482/CH4/EX4.3/Ex4_3.sce b/3482/CH4/EX4.3/Ex4_3.sce
new file mode 100644
index 000000000..3cde04b65
--- /dev/null
+++ b/3482/CH4/EX4.3/Ex4_3.sce
@@ -0,0 +1,20 @@
+clc;
+//page 168
+//Take x axis parallel to track and Y axis perpendicular to track
+W=25;//kN
+// Resolving weight
+Wx=W*cos(25*%pi/180);//kN
+Wy=-W*sin(25*%pi/180);//kN
+//At equillibrium equations are +-> sum Fx=0, +sum(M_A)=0, +sum(M_B)=0
+
+//+sum(M_A)=0 gives -(10.5kN)(625 mm)-(22.65 kN)(150 mm)+ R2(1250 mm)=0, R2 assumed to be in +ve Y direction
+R2=(10.5*625+22.65*150)/1250;//kN
+printf("R2=%.0f kN  +ve sign shows reaction is directed as assumed \n",R2);
+
+//+sum(M_B)=0 gives (10.5kN)(625 mm)-(22.65 kN)(150 mm)+ R1(1250 mm)=0, R1 assumed to be in +ve Y direction
+R1=(10.5*625-22.65*150)/1250;//kN
+printf("R1=%.1f kN  +ve sign shows reaction is directed as assumed \n",R1);
+
+//Sum Fx=0 gives, 22.65 N-T=0
+T=22.65;//kN
+printf("T=%.2f kN  +ve sign shows reaction is directed as assumed \n",T);
diff --git a/3482/CH4/EX4.4/Ex4_4.sce b/3482/CH4/EX4.4/Ex4_4.sce
new file mode 100644
index 000000000..2945fa1f3
--- /dev/null
+++ b/3482/CH4/EX4.4/Ex4_4.sce
@@ -0,0 +1,13 @@
+clc;
+////page 168
+Ax=4.5//in m
+Ay=6//in m
+DF=sqrt((Ax^2)+(Ay^2))
+F=150//in KN
+Ex=-(Ax/DF)*F
+printf("Ex=%.2f kN \n",Ex);
+Ey=((Ay/DF)*F)+(4*20)
+printf("Ey=%.2f kN \n",Ey);
+
+M_E=-((20*7.2)+(20*5.4)+(20*3.6)+(20*1.8)-((Ay/DF)*F*Ax))
+printf("M_E=%.0f kN  +ve sign shows reaction is directed as assumed \n",M_E);
diff --git a/3482/CH4/EX4.5/Ex4_5.sce b/3482/CH4/EX4.5/Ex4_5.sce
new file mode 100644
index 000000000..49ea8daca
--- /dev/null
+++ b/3482/CH4/EX4.5/Ex4_5.sce
@@ -0,0 +1,23 @@
+clc;
+//page 169
+
+//At equillibrium  +sum(Mo)=0,
+//s=r*theta;
+//F=k*s=k*r*theta;
+k=45;//N/mm
+r=75;//mm
+W=1800;//N
+l=200;//mm
+
+
+// trial and error 
+printf("Probable answers by trial and error method are \n");
+for i=0:0.1:%pi/2 // from 0 to 90 degrees
+
+difference=(sin(i)-k*r^2*(i)/(W*l));
+if difference<0.01 then   // Approximation
+   theta=i;
+   theta=theta*180/%pi;//Degrees , conversion into degrees
+printf("Theta=%.2f degrees\n",theta); 
+end
+end
diff --git a/3482/CH4/EX4.6/Ex4_6.sce b/3482/CH4/EX4.6/Ex4_6.sce
new file mode 100644
index 000000000..113c314d6
--- /dev/null
+++ b/3482/CH4/EX4.6/Ex4_6.sce
@@ -0,0 +1,33 @@
+clc;
+//page 185
+
+m=10;//kg mass of joist
+g=9.81;//m/s^2 gravitational acceleration
+W=m*g;//N
+AB=4;//m
+// Three force body
+BF=AB*cos(45*%pi/180);//m
+AF=BF;//m
+
+AE=1/2*AF;//m
+EF=AE;//m
+CD=AE;//m
+BD=CD/tan((45+25)*%pi/180);//m
+DF=BF-BD;//m
+CE=DF;//m
+alpha=atan(CE/AE);//radians
+alpha=alpha*180/%pi;//degrees
+
+//From geometry
+
+G=90-alpha;//degrees
+B=alpha-(90-(45+25));//degrees
+C=180-(G+B);//Degrees
+
+//Force triangle
+//T/sin(G)=R/sin(C)=W/sin(B)..... sine law
+
+T=W/sin(B*%pi/180)*sin(G*%pi/180);//N
+R=W/sin(B*%pi/180)*sin(C*%pi/180);//N
+printf("Tension in cable T= %.1f N\n Reaction At A is \n R= %.1f N with angle alpha= %.1f degrees with +ve X axis",T,R,alpha); 
+
diff --git a/3482/CH4/EX4.7/Ex4_7.sce b/3482/CH4/EX4.7/Ex4_7.sce
new file mode 100644
index 000000000..60de967de
--- /dev/null
+++ b/3482/CH4/EX4.7/Ex4_7.sce
@@ -0,0 +1,20 @@
+clc;
+//page 194
+m1=80;//kg mass of man
+m2=20;//kg, mass of ladder
+m=m1+m2;//kg
+g=9.81;//m/s^2 gravitational acceleration
+W=-m*g;//N, j
+
+C=-0.6*W/3;//N
+Bz=-0.6*C/1.2;//N
+By=-0.9*W/1.2;//N
+
+printf(" Reaction At B is B= (%.0f) N j +(%.1f N)k\n",By,Bz);
+printf(" Reaction At C is C= (%.2f) N k\n",C);  
+Ay=-W-By;//N
+Az=-C-Bz;//N
+
+
+printf(" Reaction At A is A= (%.0f) N j +(%.1f N)k \n",Ay,Az); 
+
diff --git a/3482/CH4/EX4.8/Ex4_8.sce b/3482/CH4/EX4.8/Ex4_8.sce
new file mode 100644
index 000000000..30856f032
--- /dev/null
+++ b/3482/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,22 @@
+clc;
+W=-1200;//N,j  Weight 
+BD=[-2.4,1.2,-2.4];//m, Vector BD
+EC=[-1.8,0.9,0.6];//m, Vector EC
+//T_BD=norm(T_BD)*BD/norm(BD);// m, vector of tension in BD
+//T_EC=norm(T_EC)*EC/norm(EC);// m, vector of tension in EC
+// Applying equillibrium conditions we get
+// Sum_F=0, and Sum(M_A)=0 and setting co-efficient equal to zero
+A=[0.8,0.771;1.6,-0.514];//MAtrix of co-efficient
+b=[-1440;0];//matrix b
+x=linsolve(A,b);// solution matrix
+T_BD=x(1);// N,Tension in BD
+T_EC=x(2);//N, Tension in EC
+printf("T_BD= (%.0f N) and T_EC= (%.0f N) \n",x(1),x(2));
+
+Ax=2/3*T_BD+6/7*T_EC;//N, x component of reaction at A
+Ay=-(1/3*T_BD+3/7*T_EC+W);//N, Y component of rection at A
+Az=2/3*T_BD-2/7*T_EC;//N, z component of reaction at A
+
+printf("Reaction at A is A=(%.0f N)i +(%.0f N)j +(%.1f N)k \n",Ax,Ay,Az);
+//Answe in Newton instead of lbs
+//1lbs=4.44N
diff --git a/3482/CH4/EX4.9/Ex4_9.sce b/3482/CH4/EX4.9/Ex4_9.sce
new file mode 100644
index 000000000..4789e095f
--- /dev/null
+++ b/3482/CH4/EX4.9/Ex4_9.sce
@@ -0,0 +1,19 @@
+clc;
+//page 198
+//Free body diagram
+m=30//in kg
+g=9.81//in m/s2
+w=-m*g//in J
+DC=[-480 240 -160]//in mm
+X=norm(DC)
+T=DC/X
+disp("Tension in the vector form=")
+disp(T)
+//Equilibrium equations
+//From equation 2, setting unit vector=0
+Ax=49//in N
+Ay=73.5//in N
+A=[Ax Ay]
+y=norm(A)
+disp("Tension in the vector form in N=")
+disp(y)