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clear
//
//Now, we cannot find a joint with only two unknown forces without finding reactions.
//Consider the equilibrium of the entire frame,Sum of moments about A is zero,Horizontal forces & Vertical forces is zero.
//variable declaration
PB=30.0 //vertical load at point B,KN
PC=50.0 //vertical load at point C,KN
PDv=40.0 //vertical load at point D,KN
PDh=20.0 //Horizontal load at point D,KN
PF=30.0 //vertical load at point F,KN
HA=PDh
RE=(PC*4+PDv*8+PDh*4+PF*4)/(8.0)
VA=PB+PC+PDv+PF-RE
//joint A
//sum of vertical & sum of horizontal forces is zero.
FAB=VA
FAF=HA
//joint E
//sum of vertical & sum of horizontal forces is zero.
FED=RE
FEF=0
//Joint B: Noting that inclined member is at 45°
//sum of vertical & sum of horizontal forces is zero.
theta=45.0
FBF=(VA-PB)/sin(theta*%pi/180)
printf("\n FBF= %0.4f KN (Tension) ",FBF)
FBC=FBF*cos(theta*%pi/180)
printf("\n FBC= %0.4f KN (Comp.) ",FBC)
//Joint C:
//sum of vertical & sum of horizontal forces is zero.
FCF=PC
printf("\n FCF= %0.4f KN (Comp.) ",FCF)
FCD=FBC
printf("\n FCD= %0.4f KN (Comp.) ",FCD)
//Joint D: Noting that inclined member is at 45°
//sum of vertical & sum of horizontal forces is zero.
theta=45.0
FDF=(RE-PDv)/cos(theta*%pi/180)
printf("\n FDF= %0.4f KN (Tensile) ",FDF)
//check
FDF=(FCD+PDh)/cos(theta*%pi/180)
printf("\n FDF= %0.4f KN Checked ",FDF)
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