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+clear
+//
+//
+
+
+//Initilization of Variables
+
+L_CB=2 //m //Length of CB
+L_AC=4 //m //Length of AB
+M_C=15 //KN.m //Moment At Pt C
+F_C=30 //KN
+L=6 //m Span of Beam
+
+//Let X=E*I
+X=10000 //KN-m**2
+
+//Calculations
+
+//Let V_A and V_B be the reactions at A & B respectively
+//V_A+V_B=30
+
+//Taking Moment a A,we get
+V_B=(F_C*L_AC+M_C)*L**-1
+V_A=30-V_B
+
+//Now Taking Moment at distacnce x from A
+//M_x=7.5*x-30*(x-4)+15
+
+//By using Macaulay's Method
+//EI*(d**2*x/dx**2)=M_x=7.5*x-30*(x-4)+15
+
+//Now Integrating above Equation we get
+//EI*(dy/dx)=C1+7.5*x**2*2**-1-15*(x-4)**2+15*(x-4)  ............(1)
+
+//Again Integrating above Equation we get
+//EIy=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1..........(2)
+
+//Boundary Cinditions
+x=0
+y=0
+
+//Substituting above equations we get 
+C2=0
+
+x=6 //m
+y=0
+
+C1=-(7.5*6**3*6**-1-5*2**3+15*2**2*2**-1)*6**-1
+
+//EIy_c=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1
+//Sub values in Above equation we get
+y_c=(93.3333*(X)**-1)
+
+//Result
+printf("\n The Deflection at C %0.4f  mm",y_c)