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+// Problem no 12.8,Page No.292
+
+clc;clear;
+close;
+
+L=8 //m //span
+W=24*10**3 //N/m //U.D.L
+y=2*10**-2 //m //deflection
+E=20*10**9
+I=10**-5 //m**4
+
+//Calculations
+
+//The Downward deflection at C Due to u.d.l
+//Y_c=5*W*L**3*(384*E*I)**-1
+
+//The Upward Deflection at C due to prop Reaction P 
+//Y_c_1=P*L**3*(48*E*I)**-1
+
+//Since the prop is at the same level as end supports
+//Y_c_1=Y_c
+P_1=5*W*8**-1*10**-3 //KN
+
+//The reaction at A and B is equal
+R_a=(24-15)*2**-1
+R_b=R_a;
+//Shear Force at B
+V_B=4.5 //KN
+
+//Shear Force at C
+V_C1=4.5-24*2**-1
+V_C2=4.5-24*2**-1+15
+
+//Shea rForce at A
+V_A=-4.5 //KN
+
+//B.M at C due to u.d.l
+M_C1=W*L*8**-1*10**-3 //KN*m
+
+//B.M due to only prop reaction P=15 KN
+P=15
+M_C2=-P*L*4**-1 //KN*m
+
+//B.M at D
+M_D=4.5*1.5-24*8**-1*1.5**2*2**-1
+
+//In second case prop sinks by 2 cm
+//Y_c-Y_c_1=2 
+
+//So Further simplifying and sunstituting values in above equation we get
+P=-(2*100**-1-(5*W*L**3*(384*E*I)**-1))*(L**3*(48*E*I)**-1)**-1
+
+//Let Each end reaction be X
+X=(24-14.625)*2**-1
+
+//Result
+printf("prop reaction is %.2f",P_1);printf(" KN")
+printf("\n The End Reaction is %.2f",X);printf(" KN")
+
+//Plotting the Shear Force Diagram
+subplot(2,1,1)
+X1=[0,4,4,8,8]
+Y1=[V_B,V_C1,V_C2,V_A,0]
+Z1=[0,0,0,0,0]
+plot(X1,Y1,X1,Z1)
+xlabel("Length x in m")
+ylabel("Shear Force in kN")
+title("the Shear Force Diagram")
+
+//Plotting the Bendimg Moment Diagram
+subplot(2,1,2)
+X2=[0,4,4]
+Y2=[0,M_C1,0]
+Z2=[0,0,0]
+plot(X2,Y2)
+xlabel("Lenght in m")
+ylabel("Bending Moment in kN.m")
+title("the Bending Moment Diagram")