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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clc;
+//Drawing of shear and bending moment diagram
+printf("Given problem is for drawing diagram, this diagram is drawn by step by step manner.\n ");
+F_AC=7200;//N/m, distributed load applied at A to C
+F_E=1800;//N, force applied at E
+AC=0.3;//m, perpendicular distance between A and B
+CD=0.15;//m, perpendicular distance between C and D
+DE=0.1;//m, perpendicular distance between E and D
+EB=0.25;//m, perpendicular distance between E and B
+AB=0.8;//m, length of beam AB
+F=F_AC*AC;//N, Force due to districuted load at AC/2
+//By free body of entire beam
+//By sum(m_A)=0
+By=(F*AC/2+F_E*(AC+CD+DE))/AB;//N,Y componet of Reaction at B
+//By sum(m_B)=0
+A=(F*(AB-AC/2)+F_E*EB)/AB;//N, Reaction at A
+//by sum(Fx)=0
+Bx=0;//N, xcomponent of rection at B
+
+//Diagrams
+//For section A to C
+//Applying sum(Fy)=0
+i=0;
+for x=0:.1:0.3
+ i=i+1;
+ X(i)=x;
+V(i)=A-F*x;//N
+//Applying sum(M1)=0
+M(i)=A*x-F/2*x^2;//N.m
+end
+
+//For section Cto D
+//Applying sum(Fy)=0
+for x=0.3:0.05:0.45
+ i=i+1;
+ X(i)=x;
+V(i)=A-F;//N
+//Applying sum(M1)=0
+M(i)=A*x-F*(x-0.15);//N.m
+end
+//For section D to B
+
+for x=0.45:.05:0.8
+
+i=i+1;
+ X(i)=x;
+ //Applying sum(Fy)=0
+ V(i)=A-F-F_E;//N
+//Applying sum(M1)=0
+M(i)=A*x-F*(x-0.15)+F_E*DE-F_E*(x-0.045);//N.m
+end
+
+
+plot(X,V,'r');//Shear diagram
+plot(X,M,'-');//Bending moment diagram