Engineering Solve X-Forces for Frame Statics Problem

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The discussion focuses on solving for X-forces in a frame statics problem, where initial calculations for Y-forces were provided. The participant calculated Dx as 300 N and Cx as 300 N, while also determining Bx to be 600 N. However, there is uncertainty regarding the correctness of the X-forces, especially since support A cannot handle horizontal loads. Feedback indicates that the value of Dx is incorrect, as member DE should be treated as a beam supported at both ends, and the forces acting on pin D must account for the constraints imposed by the frame's geometry. The conversation emphasizes the importance of correctly applying the method of sections and understanding the role of supports in determining forces.
bob1352
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Homework Statement
This is a weekly assignment problem.
Relevant Equations
Fnetx = 0
Fnety = 0
Sum of Moments at a point = 0
So for this problem I have already solved for the Y forces: Dy = 171.43 N, Cy = 228.57 N, and By = -428.57 N. For the X forces I split up the frame and took the moment of DE.

Me = 1.5(300) + 3.5(300) - 5(Dx), Dx = 300

For CD Dx = -Cx fo I got Cx = 300, as when you forces are two member Dx = -300 on member CD. Then I used the moment of ABC to solve for Bx

Ma = -2.5(Bx) + (5)300, Bx = 600 N.

I am almost postive my Y forces are correct, but I am unsure of the X-forces and I can balance the moments, but not the sum of the forces. Am I correct or am I missing something?
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Note that support A is of a type that is unable to take horizontal loads.
Link BD is working under compression, while link BE is working under tension.
Therefore, there are two x-component forces acting on B in opposite directions.
 
So Bx, Cx and Dx should be zero then? I am not using the method of joints, but the technique for multi-force members.
 
bob1352 said:
So Bx, Cx and Dx should be zero then?
What makes you to think that way?
bob1352 said:
I am not using the method of joints, but the technique for multi-force members.
Could you show us the FBD that you have done so far?
 
Lnewqban said:
What makes you to think that way?

Could you show us the FBD that you have done so far?
Sure, here you go. I split the frame into parts as what the method of sections suggests.
 

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bob1352 said:
Sure, here you go. I split the frame into parts as what the method of sections suggests.
Apologies about the delayed response.
Your work looks good, but your value of Dx is incorrect.
Member DE acts as a beam supported at both ends, not like a cantilever beam.
That is because pivot E can't provide any reactive moment.

Simultaneously, pin D is not free to rotate around pivot E under the two lateral loads.
A force must be preventing that rotation from happening.
That force reaches pin D via ground-member AB-pin B-member BD.

Frame Statics Problem (1).jpg
 

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