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+// Problem no 6.13,Page No.163
+
+clc;clear;
+close;
+
+L=6 //m //Span of Beam
+sigma=100*10**6 //Pa //Bending stress
+E=210*10**9
+y=0.45 //m //Depth
+
+//Calculations
+
+//Taking moment at B
+R_a=20*6*3+6*40*2*2**-1
+
+//At a section x from A the rate of Loading=20+2*3**-1*x //KN/m
+//S.F=100-20*x-x**2*3**-1
+//M=100*x-10*x**2-x**3*9**-1 
+
+//Thus B.M will be max where S.F is zero,we get equation as
+//x**2+60*x-300=0
+a=1
+b=60
+c=-300
+
+X=b**2-4*a*c
+x_1=(-b+X**0.5)*(2*a)**-1
+x_2=(-b-X**0.5)*(2*a)**-1
+
+x=4.641
+M=100*x-10*x**2-x**3*9**-1 //KN*m //Max bending moment
+I=M*sigma**-1*y*1000*2**-1 //m**4 //M.I
+
+//E*I*d**2*y*(d*x**2)**-1=-100*x+10*x**2+x**3*9**-1
+
+//AFter Integrating above EquATION WE get
+//E*I*dy*(dx)**-1=-50*x**2+10*3**-1*x**3+x**4*36**-1+C_1
+//Again Integrating above EquATION WE get
+//E*I*y=-50*x**3*3**-1+10*12**-1*x**4+x**5*180**-1+C_1*x+C_2
+
+//At x=0,y=0  ,C_2=0
+//At x=6,y=0 
+x=6
+C_2=0
+C_1=-(-50*x**3*3**-1+10*12**-1*x**4+x**5*180**-1)*x**-1
+
+x=3 //m
+y=1*(E*I)**-1*(-50*x**3*3**-1+10*12**-1*x**4+x**5*180**-1+C_1*x+C_2)*1000*100
+
+//Result
+printf("The central Deflection is %.2f",y);printf(" cm")