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+clear
+//
+//
+
+//Initilization of Variables
+
+L_AC=2 //m //Length of BD,CB,AC
+L_BD=2
+L_CB=2
+F_C=40 //KN //Force at C
+F_D=10 //KN Force at D
+L=6 //m spna of beam
+
+//EI is constant in this problem
+
+//Calculations
+
+//Let V_A & V_B be the reactions at A & B Respectively
+//V_A+V_B=50
+
+//Taking Moment at Pt A
+V_B=(F_D*L+F_C*L_AC)*(L_AC+L_CB)**-1
+V_A=50-V_B
+
+//Now Taking Moment at distance x from A,M_x
+//M_x=15*x-40*(x-2)+35*(x-4)
+//EI*(d**2*y/dx**2)=15*x-40*(x-2)+35*(x-4)
+
+//Now Integrating above equation we get
+//EI*(dy/dx)=C1+7.5*x**2-20*(x-2)**2+17.5(x-4)**2
+
+//Again Integrating above equation we get
+//EI*y=C2+C1*x+2.5*x**2-20*3**-1*(x-2)**3+17.5*(x-4)**3*3**-1
+
+//At
+x=0
+y=0
+//we get
+C2=0
+
+//At
+x=4
+y=0
+//we get
+C1=(2.5*4**3-20*3**-1*2**3)*4**-1
+
+//Now Deflection at C
+x=2
+C1=-26.667
+C2=0
+y_C=C2+C1*x+2.5*x**3
+
+//Now Deflection at D
+C1=-21.667
+C2=0
+y_D=-26.667*6+2.5*6**3-20*3**-1*4**3+17.5*2**3*3**-1
+
+//Result
+printf("\n Deflections Under Loads are:y_D %0.4f ",y_D)
+printf("\n :y_C %0.2f ",y_C)