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//Finding forces in all members
//Refer fig. 4.11(a)
theta1=atand(4/6) //Degree
theta2=atand(8/6) //Degree
theta3=atand(4/3) //Degree
//Joint H
FHG=20/sind(53.13) //kN (Compression)
FHF=25*cosd(53.13) //kN (Tension)
//Taking moment about A
RG=(20*9+12*6)/6 //kN
VA=32-42 //kN (Downwards)
HA=0
//Joint A
//applying equilibrium conditions
FAC=10/sind(33.69) //kN (Compression)
FAB=18.03*cosd(33.69) //kN (Tension)
//Joint B
FBC=0
FCE=FAC //kN (Compression)
//Joint D
FDE=0
FDF=FBD //kN (Tension)
//Joint E
FEF=0
FEG=FCE //kN (Compression)
//Joint F
FAG=12 //kN (Compression)
printf("Required values are:-\nFHG=%.2f kN (Compression)\nFHF=%.2f kN (Tension)\nRG=%.2f kN\nVA=%.2f kN (Downwards)\nHA=%.2f kN\nFAC=%.2f kN (Compression)\nFAB=%.2f kN (Tension)\nFBC=%.2f kN\nFCE=%.2f kN (Compression)\nFDE=%.2f kN\nFDF=%.2f kN (Tension)\nFEF=%.2f kN\nFEG=%.2f kN (Compression)\nFAG=%.2f kN (Compression)",FHG,FHF,RG,-VA,HA,FAC,FAB,FBC,FCE,FDE,FDF,FEF,FEG,FAG)
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