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Determining internal force with section cut on structure

  • Thread starter Woopydalan
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  • #1
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Hello, my problem seems to be that I have two values that are possible for the Force FG, but I also am not considering the internal forces F_DE and F_FE. I'm not sure how to go about finding those if they are important for solving. For the summation of forces in a direction, do I only consider external forces and reaction forces, and not internal forces?
 

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  • #2
PhanthomJay
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It is almost always best to find the external end reactions first before cutting sections to find the internal member forces. By so doing you can find A_y; and the value of A_x should readily pop out at you.
You then have some plus/ minus errors in your sum of moments equations....if cw is minus, then ccw is plus. Otherwise , your free body diagrams look ok. Remember that when you solve for the unknown forces, and you get a negative number, it sometimes means you assumed the wrong direction for the force. It does not necessarily imply compression. Note also that when you sum forces or moments of a cut section, you include external forces and you also treat the internal forces at the cut members as external to the cut section.
 
  • #3
nvn
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Woopydalan: PhanthomJay well covered most of your post 1 questions, so I will only mention a few additional items, as follows. I do not know what you mean by "Force FG" in post 1.

... but I also am not considering the internal forces F_DE and F_FE. I'm not sure how to go about finding those, if they are important for solving.
They are not important for the given question. You can continue excluding and ignoring them. Good call.

By the way, I currently notice the given truss seems unstable (partially constrained) to the second degree. However, to answer the given text book question, this instability issue does not seem to matter. You can still solve for, and answer, the given question, by following the advice given by PhanthomJay in post 2.
 
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  • #4
PhanthomJay
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nvn:

You are correct, joint B is unstable under the 35 kN load in BC. The truss should not be built this way with missing triangles.
 
  • #5
nvn
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Woopydalan: Hint: As PhanthomJay mentioned, find external reaction forces before cutting section cut a-a. summation(Fx) = 0 = Ax; therefore, Ax = 0 kN. Try again.
 
  • #6
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How does it look now?
 

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  • #7
nvn
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Woopydalan: Somehow, in your second free-body diagram, you inadvertently omitted a 35 kN applied load at joint C. Try again.
 
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  • #8
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Without ommiting the 35 kN, then the answer should be -75 kN and 75 kN?
 
  • #9
nvn
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Nice work, Woopydalan. Your answer is correct.
 
  • #10
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Thank you!
 

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