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+// Problem no 4.4.3,Page No.91
+
+clc;clear;
+close;
+AC=5000 //N/m //u.v.l
+L_AB=4 //m //Length of AB
+
+//Calculations
+
+//Consider a section at Distance x from B
+//DB=x
+//By similar triangles (triangle ABC and BDE) we get
+
+//Shear Force at x
+//F_x=-DB*DE*2**-1
+//After substituting values in above equation we get
+//F_x=625*x**2
+
+//shear Force at B where x=0
+V_B=0
+
+//shear Force at A where x=L_AB=4
+V_A=625*L_AB**2
+
+//Bending Moment Calculation
+
+//M_x=DB*DE*DB*3**-1*2**-1
+//Substituting values in above equation we get
+//M_x=-625*x**3*3**-1
+
+//Bending Moment at B where x=0
+M_B=0
+
+//Bending Moment at A where x=L_AB=4
+M_A=-625*L_AB**3*3**-1
+
+//Result
+printf("The Shear Force and Bending Moment Diagrams are the results")
+
+//Plotting the Shear Force Diagram
+subplot(2,1,1)
+X1=[0,L_AB]
+Y1=[V_B,V_A]
+Z1=[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)
+Y2=[M_B,M_A]
+X2=[0,L_AB]
+Z2=[0,0]
+plot(X2,Y2)
+xlabel("Length in m")
+ylabel("Bending Moment in kN.m")
+title("the Bending Moment Diagram")