Structure Analysis - Problems 3

[Dito05]

Hi guys, can anyone help me with this problem?

 

[SWComposites]

So what help do you need?
Are you a student?
Maybe post your proposed solution along with specific questions.

 

[GregLocock]

Draw 2 FBDs.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[Dito05]

No, I am not a student right now. However, I wanted to practice a few old problems from a book I have, just to keep my brain working as before, and I got stuck specifically with this one.
I tried drawing 2 FBDs but when I tried to solve the member first that has Pins at A and B when solving my moment equation located at pin A, I got confused since there is a 0.2m distance from the center of the member to the location of the pin at B.
Hence, I got a moment equation at pin A, with my distributed load and 2 reactions the By and Bx.
I don't know if I am explaining my self clearly, but I just want to know from this example what to do in scenarios such as this one.

Thanks in advance.

 

[rb1957]

since it's pinned at A and C, it is statically determinate.
Consider the applied load creates a moment at A, but A is pinned. So BC reacts the moment at A, and A reacts the nett vertical load (applied-BC component) and the horizontal component of BC.

So the horizontal couple between A and C reacts the applied moment at A. MA = 3*3*1.5, and AC = 1.5, so the horizontal couple is 9kN.
Then the vertical component of BC is 9*1.5/2, and the vertical component at A is 9 (applied)-9*1.5/2 = 9/4.

See if this makes sense, show us your work, then we can help you understand it.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[desertfox]

I’m in agreement with rb1957 in that it’s a statically determinable problem, treat the strut as a two force member, therefore the strut reaction can only act parallel to the line of action of the strut.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Ingenuity]

AC = 1.5

AC = 1.7, so horizontal reactions are 3*3*1.5/1.7 = 7.94 kN not 9 kN

 

[Dito05]

I got it guys, Thank you very much for the help!
@rb1957, @desertforx, @ingenuity thanks for the tips, they worked
I attached my work too.