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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
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