10 files changed, 343 insertions, 0 deletions
diff --git a/1205/CH7/EX7.1/S_7_1.sce b/1205/CH7/EX7.1/S_7_1.sce
new file mode 100644
index 000000000..e6c070692
--- /dev/null
+++ b/1205/CH7/EX7.1/S_7_1.sce
@@ -0,0 +1,31 @@
+clc;
+P=2400;//N, Vertical Force applied at D
+AB=2.7;//m, perpendicular distance between A and B
+BE=2.7;//m, perpendicular distance between E and B
+BK=1.5;//m, perpendicular distance between B and K
+AJ=1.2;//m, perpendicular distance between A and J
+EF=4.8;//m, perpendicular distance between E and F
+BD=3.6;//m, perpendicular distance between D and B
+//For entire truss
+//By free body diagram we get the force at A, B , c
+A=1800;//N
+B=1200;//N
+C=3600;//N
+alpha=atan(EF/(AB+BE));//rad
+//a. Internal forces at j
+//Applying sum(M_J)=0
+M=A*AJ;//N.m,Couple on member ACF at J
+//Applying sum(Fx)=0
+F=A*cos(alpha);//N, Axial force at J
+//Applying sum(Fy)=0
+V=A*sin(alpha);//N, shearing force at J
+printf("Thus, Internal forces at J are equivalent to \n Couple M = %.0f N.m \n Axial force F= %.0f N \n Shearing force V= %.0f N\n",M,F,V);
+
+//a. Internal forces at K
+//Applying sum(M_K)=0
+M=B*BK;//N.m,Couple on frame
+//Applying sum(Fx)=0
+F=0;//N, Axial force at J
+//Applying sum(Fy)=0
+V=-B;//N, shearing force at J
+printf("Thus, Internal forces at K are equivalent to \n Couple M = %.0f N.m \n Axial force F= %.0f N \n Shearing force V= %.0f N\n",M,F,V);
diff --git a/1205/CH7/EX7.10/S_7_10.sce b/1205/CH7/EX7.10/S_7_10.sce
new file mode 100644
index 000000000..2984c5dec
--- /dev/null
+++ b/1205/CH7/EX7.10/S_7_10.sce
@@ -0,0 +1,30 @@
+clc;
+AB=150;//m, distance AB
+s=30;//m, sag of cable
+w=45;//N/m Uniform weigth per unit length of cable
+
+//Equation of cable, by 7.16
+//Coordinates of B
+
+xB=AB/2;//m
+C=[99,105,98.4,90];//trial values
+
+for i=1:4
+    if ((30/C(i)+1)-cosh(xB/C(i)))<0.0001 then c=C(i);
+    break;
+end
+end
+yB=s+c;//m
+
+//Maximum and minimum values of tension
+Tmin=w*c;//N, To
+Tmax=w*yB;//N TB
+printf("Minimum value of tension in cable is Tmin= %.0f N\n",Tmin);
+printf("Maximum value of tension in cable is Tmax= %.0f N\n",Tmax);
+//Length of cable
+
+S_CB=sqrt(yB^2-c^2);//m, one halph length by 7.17
+S_AB=2*S_CB;//m, full length of cable
+
+printf("Fulllength of cable is s_AB= %.0f m\n",S_AB);
+
diff --git a/1205/CH7/EX7.2/S_7_2.sce b/1205/CH7/EX7.2/S_7_2.sce
new file mode 100644
index 000000000..795251925
--- /dev/null
+++ b/1205/CH7/EX7.2/S_7_2.sce
@@ -0,0 +1,53 @@
+clc;
+//Drawing of shear and bending moment diagram
+printf("Given problem is for drawing diagram, this diagram is drawn by step by step manner.\n ");
+F_A=-20;//kN, force applied at A
+F_C=-40;//kN, force applied at C
+AB=2.5;//m, perpendicular distance between A and B
+BC=3;//m, perpendicular distance between C and B
+CD=2;//m, perpendicular distance between C and D
+//By free body of entire beam
+//By sum(m_D)=0
+R_B=-(CD*F_C+(AB+BC+CD)*F_A)/(BC+CD);//kN, Reaction atB
+//By sum(m_A)=0
+R_D=-(BC*F_C-(AB)*F_A)/(BC+CD);//kN, Reaction atB
+//For section 1
+//Applying sum(Fy)=0
+V1=F_A;//kN
+//Applying sum(M1)=0
+M1=V1*0;//kN.m
+
+//For section 2
+//Applying sum(Fy)=0
+V2=F_A;//kN
+//Applying sum(M1)=0
+M2=F_A*AB;//kN.m
+
+//For section 3
+//Applying sum(Fy)=0
+V3=R_B+F_A;//kN
+//Applying sum(M1)=0
+M3=F_A*AB;//kN.m
+
+//For section 4
+//Applying sum(Fy)=0
+V4=R_B+F_A;//kN
+//Applying sum(M1)=0
+M4=F_A*(AB+BC)+R_B*BC //kN.m
+
+//For section 5
+//Applying sum(Fy)=0
+V5=R_B+F_A+F_C;//kN
+//Applying sum(M1)=0
+M5=F_A*(AB+BC)+R_B*BC//kN.m
+
+//For section 6
+//Applying sum(Fy)=0
+V6=R_B+F_A+F_C;//kN
+//Applying sum(M1)=0
+M6=V6*0//kN.m
+X=[0,2.5,2.5,5.5,5.5,7.5]
+V=[V1,V2,V3,V4,V5,V6];//Shear matrix
+M=[M1,M2,M3,M4,M5,M6];//Bending moment matrix
+plot(X,V);//Shear diagram
+plot(X,M,'r');//Bending moment diagram
diff --git a/1205/CH7/EX7.3/S_7_3.sce b/1205/CH7/EX7.3/S_7_3.sce
new file mode 100644
index 000000000..e8c89eef3
--- /dev/null
+++ b/1205/CH7/EX7.3/S_7_3.sce
@@ -0,0 +1,55 @@
+clc;
+//Drawing of shear and bending moment diagram
+printf("Given problem is for drawing diagram, this diagram is drawn by step by step manner.\n ");
+F_AC=7200;//N/m, distributed load applied at A to C
+F_E=1800;//N, force applied at E
+AC=0.3;//m, perpendicular distance between A and B
+CD=0.15;//m, perpendicular distance between C and D
+DE=0.1;//m, perpendicular distance between E and D
+EB=0.25;//m, perpendicular distance between E and B
+AB=0.8;//m, length of beam AB
+F=F_AC*AC;//N, Force due to districuted load at AC/2
+//By free body of entire beam
+//By sum(m_A)=0
+By=(F*AC/2+F_E*(AC+CD+DE))/AB;//N,Y componet of Reaction at B
+//By sum(m_B)=0
+A=(F*(AB-AC/2)+F_E*EB)/AB;//N, Reaction at A
+//by sum(Fx)=0
+Bx=0;//N, xcomponent of rection at B
+
+//Diagrams
+//For section A to C
+//Applying sum(Fy)=0
+i=0;
+for x=0:.1:0.3
+    i=i+1;
+    X(i)=x;
+V(i)=A-F*x;//N
+//Applying sum(M1)=0
+M(i)=A*x-F/2*x^2;//N.m
+end
+
+//For section Cto D
+//Applying sum(Fy)=0
+for x=0.3:0.05:0.45
+    i=i+1;
+    X(i)=x;
+V(i)=A-F;//N
+//Applying sum(M1)=0
+M(i)=A*x-F*(x-0.15);//N.m
+end
+//For section D to B
+
+for x=0.45:.05:0.8
+
+i=i+1;
+    X(i)=x;
+    //Applying sum(Fy)=0
+    V(i)=A-F-F_E;//N
+//Applying sum(M1)=0
+M(i)=A*x-F*(x-0.15)+F_E*DE-F_E*(x-0.045);//N.m
+end
+
+
+plot(X,V,'r');//Shear diagram
+plot(X,M,'-');//Bending moment diagram
diff --git a/1205/CH7/EX7.4/S_7_4.sce b/1205/CH7/EX7.4/S_7_4.sce
new file mode 100644
index 000000000..8f692a316
--- /dev/null
+++ b/1205/CH7/EX7.4/S_7_4.sce
@@ -0,0 +1,53 @@
+clc;
+//Drawing of shear and bending moment diagram
+printf("Given problem is for drawing diagram, this diagram is drawn by step by step manner.\n ");
+F_B=500;//N, force applied at B
+F_C=500;//N, force applied at C.
+F_DE=2400;//N/m, distributed load applied at D to E
+AB=0.4;//m, perpendicular distance between A and B
+BC=0.4;//m, perpendicular distance between C and B
+CD=0.4;//m, perpendicular distance between C and D
+DE=0.3;//m, perpendicular distance between E and D
+F_E=F_DE*DE;//N, force exerted at DE/2 from E
+
+//By free body of entire beam
+//By sum(m_D)=0
+A=(CD*F_C+(BC+CD)*F_B-F_E*DE/2)/(AB+BC+CD);//N, Reaction at A
+//By sum(Fy)=0
+Dy=F_C+F_B+F_E-A;//N,Y component of  Reaction at D
+//By sum(Fx)=0
+Dx=0;//N,Y component of  Reaction at D
+//For section 1
+//Applying sum(Fy)=0
+V1=A;//N, shear force from A to B
+
+//For section 2
+//Applying sum(Fy)=0
+V2=A-F_B;//N, shear force from B to C
+
+//For section 3
+//Applying sum(Fy)=0
+V3=A-F_B-F_C;//N, shear force  from C to D
+
+//For section 4
+//Applying sum(Fy)=0
+V4=A-F_B-F_C+Dy;//N, shear force At D
+
+//For section 5
+//Applying sum(Fy)=0
+V5=0;//N, shear force at A
+//Area under bending curve is change in bending moment of that 2 points
+MA=0;//N.m
+MB=MA+V1*AB;//N.m
+MC=MB+V2*BC;//N.m
+MD=MC+V3*CD;//N.m
+ME=MD+1/2*V4*AB;//N.m
+
+
+X=[0,0.4,0.4,0.8,0.8,1.2,1.2,1.5];
+V=[V1,V1,V2,V2,V3,V3,V4,V5];//Shear matrix,
+
+plot(X,V);//Shear diagram
+X=[0,AB,AB+BC,AB+BC+CD,AB+BC+CD+DE];
+M=[MA,MB,MC,MD,ME];//Bending moment matrix
+plot(X,M,'r');//Bending moment diagram
diff --git a/1205/CH7/EX7.5/S_7_5.sce b/1205/CH7/EX7.5/S_7_5.sce
new file mode 100644
index 000000000..c6370eb0d
--- /dev/null
+++ b/1205/CH7/EX7.5/S_7_5.sce
@@ -0,0 +1,49 @@
+clc;
+//Drawing of shear and bending moment diagram
+printf("Given problem is for drawing diagram, this diagram is drawn by step by step manner.\n ");
+
+w=20;//kN/m, distributed load applied at D to E
+AB=6;//m, perpendicular distance between A and B
+BC=3;//m, perpendicular distance between C and B
+
+F_B=w*AB;//kN, force exerted at AB/2 from A
+
+//By free body of entire beam
+//By sum(m_C)=0
+RA=(F_B*(AB/2+BC))/(AB+BC);//kN, Reaction at A
+
+//By sum(m_A)=0
+RC=(F_B*(AB/2)/(AB+BC));//kN, Reaction at C
+
+//For section 1
+//Applying sum(Fy)=0
+VA=RA;//N, shear force just to right to A
+
+//For section 2
+//Applying sum(Fy)=0
+VB=VA-F_B;//kN, shear force just left to B
+
+//For section 3
+//Applying sum(Fy)=0
+VC=VB;//kN, shear force  from B to C
+
+
+//Bending moment at each end is zero
+// Maximum bending moment is at D where V=0
+VD=0;//kN
+
+x=-(VD-VA)/w;//m, location of maximum bending moment
+printf("Maximum bending moment is at D x= %.0f m from A\n",x);
+MA=0;//kN.m
+MD=MA+1/2*VA*x;//kN.m, maximum bending moment is at D
+MB=MD+1/2*VB*(AB-x);//N.m
+MC=MB+VB*BC;//N.m
+
+printf("Maximum bending moment is at MD= %.0fkN. m from A\n",MD);
+X=[0,x,AB,AB+BC];//m, 
+V=[VA,VD,VB,VC];//kN,Shear matrix,
+
+plot(X,V);//Shear diagram
+X=[0,x,AB,AB+BC];//m
+M=[MA,MD,MB,MC];//kN.m,Bending moment matrix
+plot(X,M,'r');//Bending moment diagram
diff --git a/1205/CH7/EX7.6/S_7_6.sce b/1205/CH7/EX7.6/S_7_6.sce
new file mode 100644
index 000000000..74eb56bac
--- /dev/null
+++ b/1205/CH7/EX7.6/S_7_6.sce
@@ -0,0 +1,3 @@
+
+clc;
+printf("Given problem is theoritical and no mathematical solving required for this problem");
diff --git a/1205/CH7/EX7.7/S_7_7.sce b/1205/CH7/EX7.7/S_7_7.sce
new file mode 100644
index 000000000..74eb56bac
--- /dev/null
+++ b/1205/CH7/EX7.7/S_7_7.sce
@@ -0,0 +1,3 @@
+
+clc;
+printf("Given problem is theoritical and no mathematical solving required for this problem");
diff --git a/1205/CH7/EX7.8/S_7_8.sce b/1205/CH7/EX7.8/S_7_8.sce
new file mode 100644
index 000000000..70e1c2551
--- /dev/null
+++ b/1205/CH7/EX7.8/S_7_8.sce
@@ -0,0 +1,40 @@
+clc;
+F_B=30;//kN, Vertical Force applied at B
+F_C=60;//kN, Vertical Force applied at C
+F_D=20;//kN, Vertical Force applied at D
+AB=6;//m, perpendicular distance between A and B
+BC=3;//m, perpendicular distance between C and B
+CD=4.5;//m, perpendicular distance between c and D
+DE=4.5;//m, perpendicular distance between D and E
+AE=6;//m, vertical perpendicular distance between A and E
+AC=1.5;//m, vertical perpendicular distance between A and C
+//For entire cable
+//Sum(M_E)=0, AB*Ax-Ay*(AB+BC+CD+DE)+F_B*(BC+CD+DE)+F_C*(CD+DE)+F_D*(DE)=0
+
+//Free body ABC 
+//Sum(M_c)=0 gives -Ax*AC-Ay*(AB+BC)+F_B*BC=0
+//we get 2 equations in Ax and Ay
+A=[AB,-(AB+BC+CD+DE);-AC,-(AB+BC)];//Matrix of coeficients
+B=[-(F_B*(BC+CD+DE)+F_C*(CD+DE)+F_D*(DE));-F_B*BC]; 
+X=linsolve(A,-B);//kN, Solution matrix
+Ax=X(1);//kN, X component of reaction at A
+Ay=X(2);//kN, Y component of reaction at A
+
+
+//a. Elevation of points B and D
+//Free body AB
+//sum(M_B)=0
+yB=-Ay*AB/Ax;//m, below A
+printf("Elevation of point B is %.2f m below A\n",yB);
+//free body ABCD
+//sum(M_D)=0
+yD=(Ay*(AB+BC+CD)-F_B*(BC+CD)-F_C*CD)/Ax;//m, above A
+printf("Elevation of point D is %.2f m above A\n",yD);
+
+//Maximum slope and maximum tension
+theta=atan((AE-yD)/DE);//rad 
+Tmax=-Ax/cos(theta);//kN, maximum tension
+theta=theta/%pi*180;//degree
+
+printf("Maximum slope is theta= %.1f degree and maximum tension in the cable is Tmax= %.1f kN \n",theta,Tmax);
+
diff --git a/1205/CH7/EX7.9/S_7_9.sce b/1205/CH7/EX7.9/S_7_9.sce
new file mode 100644
index 000000000..e443b59d1
--- /dev/null
+++ b/1205/CH7/EX7.9/S_7_9.sce
@@ -0,0 +1,26 @@
+clc;
+yB=0.5;//m, sag of the cable
+m=0.75;//kg/m, mass per unit length
+g=9.81;//m/s^2, acceleration due to gravity
+AB=40;//m, distance AB
+//a. Load P
+w=m*g;//N/m , Load per unit length
+xB=AB/2;//m, distance CB
+W=w*xB;//N, applied at halfway of CB
+
+//Summing moments about B
+//sum(M_B)=0
+To=W*xB/2/yB;//N
+//from force triangle
+TB=sqrt(To^2+W^2);//N, =P, as tension on each side is same
+printf("Magnitude of load P= %.0f N \n",TB);
+//slope of cable at B
+theta=atan(W/To);//rad
+theta=theta*180/%pi;//degree, conversion to degree
+printf("Slope of cable at B is theta= %.1f degree\n",theta);
+//length of cable
+//applying eq. 7.10
+sB=xB*(1+2/3*(yB/xB)^2);//m
+
+printf("Total length of cable from A to B is Length= %.4f m\n",2*sB);
+