1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normal Stress: Compression vs Tension

  1. Jan 8, 2017 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    Equation of Equilibrium (Horizontal and Vertical Forces, Moments)
    Normal Stress = F/A

    3. The attempt at a solution
    I have already solved the solution for this problem. For part (a), I simply found the force in the link, and used the cross area where the pin was, since I assumed that was where the largest normal force would be. It turned out the the link was in tension. However when I tried part (b), I got the wrong answer, and later found out it was because the cross-sectional area changed due the fact that in this case, the link was under compression. The cross-sectional area ended up being the middle of the link (not where the pin was). Why is this so? I cannot find where the book mentions this, and I cannot find any information on it with google.
  2. jcsd
  3. Jan 8, 2017 #2
    The fact that the link is in tension for theta = 0 and in compression is fairly obvious. OP, are you OK with that part?

    Now, for both cases, the link BD is a two force member, which means that the forces is axial (along the link) and is constant at all locations (neglecting end effects right at the pin holes). Thus it really does not appear to make sense to say that you used the cross sectional area at any particular location; it is a constant along the length. To speak of the "the cross sectional area ended up being the middle of the link" simply does not make sense.

    The forces values involved with likely be different for the two cases, but the area factor should be exactly the same.
  4. Jan 8, 2017 #3
    When I look at the solutions to these problems, I keep seeing that when a two force member is in tension, they use A = t(w-d) , as depicted by this image:
    When the member is in compression, they just use t*w, which to me (correct me if I'm wrong) means anywhere else along the member where there are no pins located.
  5. Jan 8, 2017 #4
    In a sense they are correct, although you did not indicate that this much dimensional data was available.

    In tension, the stress in the main body of the link is still based on the area A = w*t, but at the pins, the area is reduced to A = (w-d)*t.

    In compression, the stress is spread over the full area, A = w*t and the two smaller areas shown are not under load at all.

    For most purposes, this is splitting hairs, but not always. To get real detailed about it, you need to look at the contact stresses between the pin and the surface of the hole in the member and the way this stress spreads out over the cross section. Even in the tension case, it is only the average stress that is based on A = (w-d)*t, and the local stress may be higher very near the contact with the pin.
  6. Jan 8, 2017 #5
    Oh okay, that makes sense. I did include the entire problem, question, diagram and all, so that is all I had. Therefore they must have expected me to assume these things, although I honestly could not find it anywhere in the section.

    Essential, what I get out of your first explanation is that stress occurs throughout the whole link. The problem asks for maximum stress however, so the second explanation does touch on that if I'm reading it correctly. What you're second post is saying essentially is that in detail and reality, there are points along that link that truly do experience non-uniform stress relative to the whole link. Essentially, in tension, the maximum normal stress will be where the area is the smallest. In compression, you say that the stress is spread over the full area.

    So is it true that the maximum stress when any member is under compression is where the area is largest? For example, lets suppose we had this exact same link, but maybe the lower half was slightly wider. Would I choose to analyze the cross-section where the width is largest?
  7. Jan 8, 2017 #6
    In either tension or compression, the stress out in the main body of the link and well removed from the pins is uniform. This is St. Venant's Principle.

    The nonuniformity happens only near the pin holes, and there it may be considerable. Without bringing in the Hertz theory of contact stresses, you will not be able to get an accurate value for the true maximum stress near the hole. The value based on the area A = (w-d)*t is simply the average over a reduced area which is, as expected, greater than the average over the larger area.
  8. Jan 8, 2017 #7
    I missed the last part of your previous post.

    No, the maximum stress occurs where the area is the smallest, provided we are speaking only of average stress calculated as P/A. The load is constant in this problem (neglecting the weight of the member), so the smaller the denominator, the larger the stress value becomes.
  9. Jan 8, 2017 #8
    Okay, so simply put, in a problem of this sort, if the member is in compression, take the cross-section where the pin is not?
  10. Jan 8, 2017 #9
    I would not try to make such a rule. Instead, for every problem, think about the basics and work from there. If you are trying to maximize a ratio of P/A, then look for large P and small A at the same location. But most of all, think about the particular situation.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Normal Stress: Compression vs Tension