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+// Problem no 4.4.8,Page No.97
+
+clc;clear;
+close;
+F_C=150 //KN //Pt LOad at C
+w=300 //KN //u.v.l
+L=6 //m //Length of beam
+L_AE=1;L_DC=2;L_CB=1;L_CD=1 //m //Lengthof AE,DC,CB
+L_ED=3 //m //Length of ED
+L_Ed=2 //m
+L_dD=1 //m
+
+//Calculations
+
+//Let R_A & R_B be the reactions at A & B
+//R_A+R_B=450
+
+//Taking Moment at A
+//M_A=0=R_B*L-F_C*(L_CD+L_ED+L_AE)-w*(2*3**-1*L_ED+L_AE)
+R_B=(F_C*(L_DC+L_ED+L_AE)+w*(2*3**-1*L_ED+L_AE))*L**-1
+R_A=450-R_B
+
+//Shear Force Calculations
+
+//Shear Force at B
+V_B=R_B
+
+//Shear Force at C
+V_C1=R_B
+V_C2=R_B-F_C
+
+//Shear Force at D
+V_D=V_C2
+
+//Shear Force at E
+V_E=V_D-w
+
+//Shear Force at A
+V_A=V_E
+
+//Pt of contraflexure
+//Let F be the pt and EF=x
+//Let w1 be the rate of Loading at D we get
+w1=w*2*3**-1
+//The rate of Loading at distance x is200*x*3**-1
+
+//V_F=0=-R_B+200*x*3**-1*x*2**-1
+//After substituting values and simplifying further we get
+L_EF=(R_A*3*100**-1)**0.5
+x=(R_A*3*100**-1)**0.5;
+//Bending Moment Calculations
+
+//Bending Moment at B
+M_B=0
+
+//Bending Moment at C
+M_C=R_B*L_CB
+
+//Bending Moment at D
+M_D=R_B*(L_CB+L_DC)-F_C*L_DC
+
+//Bending Moment at E
+M_E=R_B*(L_CB+L_DC+L_ED)-F_C*(L_DC+L_ED)-w*L_Ed
+
+//Bending Moment at A
+M_A=0
+
+//Bending Moment at F
+M_F=R_A*(L_AE+L_EF)-200*x*3**-1*x*2**-1*x*3**-1
+
+L_FD=L_ED-L_EF
+
+
+//Result
+printf("The Shear Force and Bending Moment Diagrams are the results")
+
+//Plotting the Shear Force Diagram
+subplot(2,1,1)
+X1=[0,L_CB,L_CB,L_CB+L_CD,L_CB+L_CD+L_ED,L_CB+L_CD+L_ED+L_AE,L_CB+L_CD+L_ED+L_AE]                   
+Y1=[V_B,V_C1,V_C2,V_D,V_E,V_A,0]
+Z1=[0,0,0,0,0,0,0]
+plot(X1,Y1,X1,Z1)
+xlabel("Length x in m")
+ylabel("Shear Force in kN")
+title("the Shear Force Diagram")
+
+//Plotting the Bending Moment Diagram
+subplot(2,1,2)
+X2=[0,L_CB,L_CB+L_DC,L_FD+L_DC+L_CB,L_CB+L_DC+L_ED,L_CB+L_DC+L_ED+L_AE]
+Y2=[M_B,M_C,M_D,M_F,M_E,M_A]          
+Z2=[0,0,0,0,0,0]
+plot(X2,Y2)
+xlabel("Length in m")
+ylabel("Bending Moment in kN.m")
+title("the Bending Moment Diagram")