Determine member forces in the attached - Answers: FAB=0FBC=0FDE=20kN cFEF=20kN cFAF=70kN cFBE=40kN...

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Determine member forces in the attached - Answers: FAB=0FBC=0FDE=20kN cFEF=20kN cFAF=70kN cFBE=40kN cFCD=40kN cFAE=0FFB=28.
	
Document Preview: Answers: FAB=0	FBC=0	FDE=20kN c	FEF=20kN c	FAF=70kN c	FBE=40kN c	FCD=40kN cFAE=0	FFB=28.3kN t 	FBD=28.3kN t	FEC=0Answers: FAB=21.3kN c	FBC=5.33kN c	FDE=0	FEF=5.33kN t	FAF=0	FBE=4kN c	FCD=0FAE=0	FFB=20kN t 	FBD=0kN	FEC=6.67kN tProblem 3.0Note that the bracing diagonals are not joined where they cross, and the diagonal members are not capable of carrying compression forces.Answers: Reactions: AY = +30kN upwards; BY = 30kN downwards; BX = 40kN to the right.Member forces: FAD =30kN compression; FAC = 0kN; FDB = 50kN tension; FDC =40kN compression; FBC =30kN tension

Robert · Accepted Answer

Solution:  
Support reactions:
∑                              
∑
Ɵ = 45 degree 
At joint C compressive force is expected in CE which is not allowed. Hence FCE = 0 
Then for horizontal equilibrium of  joint C, FBC = 0 
For vertical equilibrium of joint C, RC = FCD = 40 kN  (C)  
Consider horizontal equilibrium of joint D.
Consider vertical equilibrium of joint D.
At joint A compressive force is expected in AE which is not allowed. Hence FAE = 0 
Then for horizontal equilibrium of  joint A, FAB = 0 
For vertical equilibrium of joint A, Ra = FAF = 70 kN  (C)  
Consider horizontal equilibrium of joint F.
Consider vertical equilibrium of joint D.

Determine member forces in the attached - Answers: FAB=0FBC=0FDE=20kN cFEF=20kN cFAF=70kN cFBE=40kN cFCD=40kN cFAE=0FFB=28. Document Preview: Answers: FAB=0 FBC=0 FDE=20kN c FEF=20kN c FAF=70kN...

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