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+// Problem no 12.3,Page No.287
+
+clc;clear;
+close;
+
+L=6 //m //span of beam
+w=30*10**3 //KN/m //u.d.l
+P=160*10**3 //N //concentrated Load
+
+//Calculations
+
+//Consider a section at a distance x from the fixed end A and B.M at x
+//M_x=R_b*(6-x)-30*2**-1*(6-x)**2-160*(3-x)
+
+//E*I*d**2y*(dx**2)**-1=-M_x=-R_b*(6-x)+15*(6-x)+160*(3-x)
+
+//Now Integrating above term we get
+//E*I*dy*(dx)**-1=R_b*2**-1*(6-x)**2-5*(6-x)**3-80*(3-x)**2+C_1  (Equation 1)
+
+//Now on Integrating we get
+//E*I*y=-R_b*6**-1*(6-x)**3+5*4**-1*(6-x)**2+80*3**-1*(3-x)**3+C_1*x+C_2  (Equation 2)
+
+//At x=0,dy*dx**-1=0
+//substituting in equation 1 we get
+//C_1=1800-R_b
+
+//At x=0,y=0
+//substituting in equation 2 we get
+//C_2=36*R_b-2340
+
+//At x=6,y=0
+R_b=72**-1*(10800-2340)
+
+//At x=0
+x=0
+M_x=R_b*(6-x)-30*2**-1*(6-x)**2-160*(3-x)
+
+//Result
+printf("Bending Moment at A is %.2f",M_x);printf(" KNm")
+printf("\n The Reaction at B %.2f",R_b);printf(" KN")