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+clear
+//
+//
+
+
+//Initilization of Variables
+L_AC=2 //m //Length of AC,CD,DB
+L_DB=2
+L_CD=2
+F_C=40 //KN //Force at C
+w=20 //KN/m //u.d.l
+L=6 //m //span of beam
+
+//Let E*I=X
+X=15000 //KN-m**2
+
+
+//Calculations
+
+//Let V_A & V_B be the reactions at A & B respectively
+//V_A+V_B=80
+
+//Taking Moment B,M_B
+V_A=(F_C*(L_CD+L_DB)+w*L_DB*L_DB*2**-1)*L**-1 //KN
+V_B=80-V_A //KN
+
+//Taking Moment at distance x from A
+//M_x=33.333*x-40*(x-2)-20*(x-4)**2*2**-1
+//EI*(d**2/dx**2)=33.333*x-40*(x-2)-10*(x-4)**2
+
+//Integrating above equation we get
+//EI*(dy/dx)=C1+33.333*x**2*2**-1-20*(x-2)**2-10*3**-1*(x-4)**3
+
+//Again Integrating above equation we get
+//EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4
+
+//At
+x=0
+y=0
+C2=0
+
+//At
+x=6
+y=0
+C1=-760*6**-1
+
+//Assuming Deflection to be max in portion CD and sustituting value of C1 in equation of slope we get
+//EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4
+//0=-126.667+33.333*x**2**-1-20*(x-2)**2
+
+//After rearranging and simplifying further we get
+
+//x**2-24*x+62=0
+//From above equations
+a=1
+b=-24
+c=62
+
+y=(b**2-4*a*c)**0.5
+
+x1=(-b+y)*(2*a)**-1
+x2=(-b-y)*(2*a)**-1
+
+//Taking x2 into account
+x=2.945 //m
+C1=-126.667
+C2=0
+
+y_max=(C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3)*X**-1 //mm
+
+//Max slope occurs at the ends
+//At A,
+//EI*(dy/dx)_A=-126.667
+//At B
+//EI*(dy/dx)_B=126.667+33.333*6**2*2**-1-20*4**2-10*2**3
+//After simplifying Further we get
+//EI*(dy/dx)_B=73.3273
+
+//Now Max slope is EI(dy/dx)_A=-126.667
+//15000*(dy/dx)_=-126.667
+
+//Let Y=dy/dx
+Y=-126.667*X**-1 //Radians
+
+//Result
+printf("\n Maximum Deflection for Beam is %0.4f  mm",y_max)
+printf("\n Maximum Slope for beam is %0.4f  radians",Y)