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clear
//
// Since all members are 3 m long, all triangles are equilateral and hence all inclined members are at 60° to horizontal. Joint-by-joint analysis is carried out . Then nature of the force is determined.
//variable declaration
AB=3.0
BC=AB
AC=AB
BD=BC
CD=BD
CE=CD
DE=CE
EF=DE
DF=DE
EG=DE
FG=DF
theta=60.0*%pi/180 //angles BAC,BCA,DCE,DEC,FEG,FGE,°
PB=40.0 //Vertical Loading at point B,KN
PD=30.0 //Vertical Loading at point D,KN
HF=10.0 //Horizontal Loading at point F,KN
PG=20.0 //Vertical Loading at point G,KN
//joint G
//sum of all vertical forces & sum of all horizotal forces is zero
FGF=PG/sin(theta)
printf("\n FGF= %0.4f KN (Tension)",FGF)
FGE=FGF*cos(theta)
printf("\n FGE= %0.4f KN (Comp.)",FGE)
//joint F
//sum of all vertical forces & sum of all horizotal forces is zero
FFG=FGF
printf("\n FFG= %0.4f KN (Comp.)",FFG)
FFE=FGF
FFD=FGF*cos(theta)+FFE*cos(theta)-HF
printf("\n FFD= %0.4f KN (Tension)",FFD)
//Now, without finding reaction we cannot proceed. Hence, consider equilibrium of the entire truss
//moment about point A
RE=((PB*AC/2)-(HF*EF*sin(theta))+(PD*(AC+CE/2))+(PG*(AC+CE+EG)))/(AC+CE)
VA=PB+PD+PG-RE
HA=HF
//joint A
//sum of all vertical forces & sum of all horizotal forces is zero
FAB=VA/sin(theta)
printf("\n FAB= %0.4f KN (Comp.)",FAB)
FAC=FAB*cos(theta)-HF
printf("\n FAC= %0.4f KN (Tension)",FAC)
//joint B
//sum of all vertical forces & sum of all horizotal forces is zero
FBC=(PB-FAB*sin(theta))/sin(theta)
printf("\n FBC= %0.4f KN (Comp.)",FBC)
FBA=FAB
FBD=-FBC*cos(theta)+FBA*cos(theta)
printf("\n FBD= %0.4f KN (Comp.)",FBD)
//joint C
//sum of all vertical forces & sum of all horizotal forces is zero
FCD=FBC*sin(theta)/sin(theta)
printf("\n FCD= %0.4f KN (Tension)",FCD)
FCE=FCD*cos(theta)+FBC*cos(theta)-FAC
printf("\n FCE= %0.4f KN (Comp.)",FCE)
//joint D
//sum of all vertical forces & sum of all horizotal forces is zero
FDE=(PD+FCD*sin(theta))/sin(theta)
printf("\n FDE= %0.4f KN (Comp.)",FDE)
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