Solving Probematic Statics: Determining Force in CD

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SUMMARY

This discussion focuses on solving statics problems, specifically determining the force in member CD using free body diagrams (FBDs). The user successfully calculated the external support reactions and the force in member BD but questioned the validity of their approach. A respondent clarified that member CD is a two-force member and suggested balancing the free body of member ABC to find the force in CD directly, emphasizing the importance of applying the 10,000 lb load at joint C when summing moments.

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I'm trying to solve the problems on this page:

http://physics.uwstout.edu/StatStr/statics/StatII/statp21f.htm

The second last one is a bit hard for me, but I think I've got it. Just need to get my approach confirmed as correct. I first found the external support reactions, that went fine.

Then, to determine the force in CD, I first drew an FBD (free body diagram) of member ABC placing two uknown forces at point B (Bx and By) and the one given force at C (should I have placed two unknown forces here also, Cx and Cy ?). Solving this I found the values for Bx and By.

I now drew a new FBD for member BD, placing two new unknown forces (Dx and Dy) in point D along with the given force in the middle, the known force E, and Bx and By (which I found earlier) in point B. Solving for Dx I got the correct answer of 9000 lb, but is my approach valid, or was this just coincidence?

Any help will be appreciated...
 
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Which problem is this? the third?
 
No, fourth.
 
Then, to determine the force in CD, I first drew an FBD (free body diagram) of member ABC placing two uknown forces at point B (Bx and By) and the one given force at C (should I have placed two unknown forces here also, Cx and Cy ?). Solving this I found the values for Bx and By.

Yes, member ABC ha Cx and Cy on its FBD.

I now drew a new FBD for member BD, placing two new unknown forces (Dx and Dy) in point D along with the given force in the middle, the known force E, and Bx and By (which I found earlier) in point B. Solving for Dx I got the correct answer of 9000 lb, but is my approach valid, or was this just coincidence?

So you took member BD, and put the forces at B (Bx and By), the forces at D (Dx and Dy, 4000 and the E force)??, plus the 5000 force in the middle??
 
You're doing too much work. You can solve for CD by just balancing the free body of member ABC now that you have the reactions at A. CD is a 2 force force member so it only has axial force (in its X direction), but you still have to apply the 10,000lb load to joint C in the Y direction when balancing the free body of ABC. Correctly sum moments on the free body of ABC and the force in CD can be found.
 

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