updated the code

1 files changed, 51 insertions, 41 deletions

diff --git a/3137/CH8/EX8.3/Ex8_3.sce b/3137/CH8/EX8.3/Ex8_3.sce
index 35a729c18..3c19ed4c0 100755
--- a/3137/CH8/EX8.3/Ex8_3.sce
+++ b/3137/CH8/EX8.3/Ex8_3.sce
@@ -1,41 +1,51 @@
-//Initilization of variables

-w=196 //N/m

-M_app=4000 //N.m

-L=6 //m

-//Calculations

-//Taking Moment about Point L and equating it to 0

-R_r=(M_app+w*L*L*0.5)/(3*L) //N

-//Taking Moment about Point R and equating it to 0

-R_l= ((((2*L)+(L/2))*(w*L))-(M_app))/(3*L) //N

-//finding point of zero shear

-a=R_l/w

-//defining x

-x0=[0,18]

-x=[0,0.5,1,1.5,2,2.5,3,3.5,a,4,4.5,5,5.5,6] //for 0<x<6

-x1=[6,12] //for6<x<12

-x2=[12,18] //for 12<x<18

-xv=[6,12,18] //specially for shear force

-xo=[12.001,12.002] //Straight line plot

-//Shear Force Calculations

-//Summing forces in vertical direction and equating to 0

-V1=R_l-w*x //N for 0<x<6

-V2=R_l-w*L //N for 6<x<18

-//Bending Moment Calculations

-M1=R_l*x-(w*x^2*0.5) //N.m for 0<x<6

-M2=R_l*x1-((w*L)*(x1-3)) //N.m for 6<x<12

-M3=R_l*x2-((w*L)*(x2-3))+M_app //N.m for 12<x<18

-Mo=[-1464.8652,2509.3333]

-//Maximum bending moment

-M_max=R_l*a*0.5 //N.m

-//Plotting

-subplot(221)

-plot(x,V1,xv,V2,x0,0)

-xtitle('Shear Force Diagram',"Span","Shear Force")

-subplot(222)

-plot(x,M1,x1,M2,x2,M3,x0,0,xo,Mo)

-xtitle('Bending Moment Diagram',"Span","Bending Moment")

-//Result

-clc

-printf('The value of reactions are R_l=%fN and R_r=%fN\n',R_l,R_r)

-printf('The point of maximum bending moment is %f meters from left support nad maximum bending moment is %fN-m\n',a,M_max)

-printf('The bending moment and shear force diagrams have been plotted')

+//Initilization of variables
+w=196 //N/m
+M_app=4000 //N.m
+L=6 //m
+//Calculations
+//Taking Moment about Point L and equating it to 0
+R_r=(M_app+w*L*L*0.5)/(3*L) //N
+//Taking Moment about Point R and equating it to 0
+R_l= ((((2*L)+(L/2))*(w*L))-(M_app))/(3*L) //N
+//finding point of zero shear
+a=R_l/w
+//defining x
+x0=[0,18]
+x=[0,0.5,1,1.5,2,2.5,3,3.5,a,4,4.5,5,5.5,6] //for 0<x<6
+x1=[6,12] //for6<x<12
+x2=[12,18] //for 12<x<18
+xv=[6,12,18] //specially for shear force
+xo=[12.001,12.002] //Straight line plot
+//Shear Force Calculations
+//Summing forces in vertical direction and equating to 0
+V1=R_l-w*x //N for 0<x<6
+V2=R_l-w*L //N for 6<x<18
+//Bending Moment Calculations
+M1=R_l*x-(w*x^2*0.5) //N.m for 0<x<6
+M2=R_l*x1-((w*L)*(x1-3)) //N.m for 6<x<12
+M3=R_l*x2-((w*L)*(x2-3))+M_app //N.m for 12<x<18
+Mo=[-1464.8652,2509.3333]
+//Maximum bending moment
+M_max=R_l*a*0.5 //N.m
+//Plotting
+subplot(221)
+plot(x,V1)
+mtlb_hold on
+plot(xv,V2*ones(length(xv),1))
+plot(x0,zeros(length(x0),1))
+mtlb_hold off
+xtitle('Shear Force Diagram',"Span","Shear Force")
+subplot(222)
+plot(x,M1)
+mtlb_hold on
+plot(x1,M2)
+plot(x2,M3)
+plot(x0,zeros(length(x0),1))
+plot(xo,Mo)
+mtlb_hold off
+xtitle('Bending Moment Diagram',"Span","Bending Moment")
+//Result
+
+printf('The value of reactions are R_l=%fN and R_r=%fN\n',R_l,R_r)
+printf('The point of maximum bending moment is %f meters from left support nad maximum bending moment is %fN-m\n',a,M_max)
+printf('The bending moment and shear force diagrams have been plotted')
\ No newline at end of file