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Cuts when finding internal forces

  1. Oct 15, 2013 #1

    I do not understand how to know where to ''cut'' the sections when finding internal forces in a structure, for example why they cut it into 3 sections at the locations the textbook did in figure 9.17.

    Thank you

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  2. jcsd
  3. Oct 16, 2013 #2
    In the figure, they're just trying to find the deflections due to each section of the rod, so they cut it in those sections to find out what the forces are. The deflections will change due to the locations of the various forces, so they cut in between them and study each section independently.
  4. Oct 16, 2013 #3
    Why would the deflection change anywhere along the bar ABC? I would think you just need to cut 2 sections, one for ABC and another for CD. I think that the internal forces are the same all along ABC, or am I mistaken?
  5. Oct 16, 2013 #4


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    I hope you meant "strain" or "stress", not "deflection". The deflection is different at every point along the bar.

    There is an applied load at point B. Why do you think that can be ignored?
  6. Oct 16, 2013 #5
    Well the force can be translated up and down, so I was thinking the point where you put it is not so relevant, meaning forces can translate
  7. Oct 16, 2013 #6


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    You need to review the difference between free vectors and bound vectors.

    And always check your ideas against common sense. If you fix an elastic band at one end and (a) pull it at the other end, or (b) pull it at its mid point, is the result of (a) and (b) the same, or different?
  8. Oct 16, 2013 #7
    Thank you for the example. I've been doing problems only with steel and brass, so I had no intuition from where the point of application would affect the stretching since I can't even notice the stretching, but the rubber band example makes perfect sense and reaches my intuition!

    Also, I had never heard of a ''bound vector'' before, thanks for mentioning it.

    However, my original question about where to apply cuts is still not clear to me!!
    Last edited: Oct 16, 2013
  9. Oct 17, 2013 #8
    So think about it: if you have a rod in uniaxial tension, you're going to experience a deflection along a certain section. In the figure, no matter where you cut in section CD, the FBD for that section will have the same force, P3. The deflection equation is d = PL/AE, so since force is constant in those sections (you can check by doing multiple FBDs), the deflection is dependent on the length. You can get the deflection for CD by d_CD = (P3)(L_CD)/AE. The same argument works for BC: the force P2 will be constant along that section). You have to take the force at C into account when you solve for P2. The deflection for d_BC = (P2)(L_BC)/AE. Again, for section AB, the force P1 will be constant in that section, so d_AB = (P1)(L_AB)/AE. So, the total deflection is d = d_AB + d_BC + d_CD.

    For this problem, the key is to realize that forces will be constant in certain sections, so we cut to determine the forces in those sections. For other problems, you will have to determine what you can take advantage of in order to know how to cut it.
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